Let's get this ball rolling.
Basic deductions:
- Any cell with a 1 or 2 is shaded.
- Any cell with a number that has a prime factor larger than the side of the grid in not shaded.
- Any cell which is adjacent to two shaded/unshaded regions, which have different numbers, must be unshaded/shaded.
- Any cell which is adjacent to a shaded/unshaded region, and if shaded/unshaded, would make that region bigger than its size, must be unshaded/shaded.
Puzzle #1:
- Numbers are distinct; their sum is 100. Every cell is part of a numbered region.
- (guessing) It seems that 28 must be unshaded and the other numbers shaded, or there will be some sort of violation (can't prove this part).
- The 24 can be 3x8 or 4x6. In any case, R1C8-R3C10 are shaded.
- The 12 can be 2x6 or 3x4. In any case, R1C1-R2C2 are shaded.
- The 20 can be 2x10 or 4x5. In any case, R9C9-R10C10 are shaded.
- If 24 is a 3x8, it will border the 12 or 20; it must be 4x6; R1C7-R4C10 are shaded.
- If the 12 is a 2x6, is will border the 16 or 24; it is 3x4; R1C1-R3C3 are shaded.
- If the 24 is R1C7-R6C10, it forces the 20 to be R9C1-R10C10, and the 16 can not be completed; the 24 is R1C5-R4C10.
- The 12 is forced to be R1C1-R4C3.
- R5C4 is shaded, or the 28 isn't big enough.
Puzzle #2:
Following those basic deductions leads to (cb-2a.png)
- R1C8, R3C10, R5C10, R9C10 must be unshaded, or an unshaded region will be a rectangle.
- R2C8, R4C10, R10C6 must be shaded; it's the only way to get 2 cells for those regions.
- R2C7, R3C8, R9C6, R10C5 must not be shaded, or some shaded region would have too many cells.
- R6C10, R7C10, R8C9 must not be shaded; it's the only way to get the 4/10 region big enough.
(cb-2b.png) 
- R1C7 must not be shaded, or an unshaded region would be a rectangle.
- R4C7 must be shaded, or the region would have too many cells.
- R7C9, R9C10 must be shaded; it's the only way to get 2 cells for that region.
- R7C8 must not be shaded, or a region would have too many cells.
- R6C7 must be shaded, or that region would have two different numbers.
(cb-2c.png)
- However the regions for R4C7 and R6C7 are placed, R4C6, R6C6, R7C6, R7C7 are all shaded.
- R8C7 is not shaded, or the 13 region isn't big enough.
- The only way to do the 6 at R6C7 is shade R6C5, R7C5, unshade R5C5, R6C4, R7C5, R7C6.
(cb2-d.png)
- R8C4 and R9C5 are shaded, or the 13 is too big.
- R5C4 is shaded or the 3 is too big. R4C5 and R5C5 are shaded; that's the only way to do the 3. R2C4, R3C3, R3C5, R4C3, R4C5 are not shaded, or 3 is too big.
- The only way to do the 4 at R4C7 is shade R3C6, R3C7, unshade R2C6.
- The only way to do the 2 at R1C6 is shade R1C5, unshade R1C4, R2C5.
- The only way to do the 3 at R10C4 is unshade R9C4, R10C4, shade R9C3, R10C3.
(cb-2e.png)
- R2C3 is shaded or it is too big. The only way to finish it is shade R1C1, R1C2, R1C3, R2C1, R2C2; unshade R3C1, R3C3.
- The only way to do R10C3 is shade R9C1, R9C2, R10C1, R10C2; unshade R8C1, R8C2, R8C3.
(cb-2f.png)
At this point there are 19 cells that have not been assigned to a completed region, and numbers 2 and 17 are not assigned to completed regions; every unassigned cell must belong to one of them. One of R4C2 and R5C1 is shaded and the other unshaded; all others must not be shaded.
(cb-2g.png)
#3:
- The 1s and 2s are shaded; some unshaded cells are forced.
- However the 2 is positioned, R3C3 is unshaded.
- If R3C2 were unshaded, that would force R4C1-R4C3 shaded, but that couldn't be expanded to a rectangle of 4; R3C2 is shaded.
- Completion of 2 and 4s is forced.
- The 8 is unshaded, or it is forced to a nonrectangular shape.
- R4C5 is shaded (adjacent to unshaded)
- R5C3 is unshaded (or would make nonrectangular shaded area)
- R1C5-R3C5 are unshaded (borders a shaded area)
- R5C4 is unshaded (only way to extend the 8)
- R6C3 is shaded (borders the 8)
- R6C1, R6C2 unshaded (borders completed region)
- However R4C5 and R6C3 go, R5C6 and R7C4 are unshaded.
- However the 8 is extended, it will border R6C5, so that cell is shaded.
- If R6C5 is not connected to either 3, then its borders force too many unshaded cells to be part of the 8; R6C6 and R7C5 are unshaded.
- Whichever way it does get connected quickly forces the other 3, the rest of the 8, and two 1x2 regions. After that, there are exactly 65 unclaimed cells left, just enough.
#4:
All the 60s are part of the same region; the other numbers are too far apart for any region to have more than one number. The sum of all these areas is exactly 100, so every region has a number.
The 60s must be unshaded; using basic logic gives:
If the 9 or 10 is unshaded, there must be some shaded cells to separate it from the 60, but there are no numbered cells available; they must be shaded.
After that, it is impossible to tell which 3 cells completes the 2, but the rest is forced.
#5:
The 2s are too far apart to be in the same region; the 61s are all a single region; other numbers are distinct. Total areas is 98; there must be two shaded squares without numbers. That's not enough to separate the 8 or 20 from the 61s; 8 and 20 must be shaded. There's only one way to place the 8 rectangle that wouldn't border another shaded region.
There are three ways to position the 20, but only one of them leaves enough space to shade two un-numbered cells. 
#6:
Basic deductions: 1s and 2s are shaded; 11 is unshaded (large prime); 7 is unshaded (the only rectangle would border a 2); 5 and R6C9 is shaded (borders the 7); R5C9 is unshaded (borders the 5); R9C2 is unshaded (borders the 2); 9 is shaded (borders the 4).
There's only one way to put rectangles that cover the 5, shaded 8, and 9, that don't border other shaded regions. R3C2 and R6C6 are unshaded, or a 2 can't be done. R5C6 is shaded, or it's too big; however it's placed, R4C6 is shaded. R6C7 is shaded, or the 7 is too big.
R10C8 is unshaded, or there's an unshaded rectangle. That completes the 7, so bordering cells are shaded. Those shaded cells form a rectangle that can't be extended, so bordering cells are unshaded. R9C4 is shaded, or the 4 is too big, completing the 2. Keep using standard logic to:
#7: There must be at least 2 regions of 4, and at least 2 regions of 9. That accounts for 99 units of area; numbers must be shared as much as possible.
- Putting R7C3 and R10C3 into a single region forces it to be a rectangle; shaded those cells and unshade the border cells.
- The 5 can't be shaded (no rectangle fits)
- R3C7 and R4C8 must be shaded (border the 5); there's only one way to fill those as rectangles.
- The 14 and 36 are unshaded (can't be in any rectangle); the 12 is shaded (borders the 14); there's only one way to do the 12 as a rectangle.
- R5C4 is shaded, or the 14 will touch the 36.
- The 6s are shaded (or they border the 36); there's only one rectangle.
- There's just enough unclaimed cells to do the 14.
#8:
The 2s are too far apart for two to be in the same region; all other numbers are distinct; adding areas is a minimum 100; every region has a number. Every number other than 2 is too big to be in a rectangle; they are unshaded. Since every cell is in a numbered region, all cells not adjacent to a 2 are unshaded.
- R1C4 is shaded, or the 34 and 16 border each other. That completes a 2.
- R3C8 is unshaded, or R4C7 can't be completed.
- R4C8 is shaded or the 16 and 5 will border each other.
- R5C9 is unshaded, or the 5 can't be completed.
- Following basic logic (region is completed, forcing borders / leaving a cell unshaded would join two regions) forces most of the rest.
#9:
- The distinct numbers sum to 99. The 32s are in the same region, and there is a single un-numbered region (1 cell, shaded).
- The 32s must be unshaded (no rectangle can hold both)
- Every other numbered region is shaded (either they border the 32, or it would take too many shaded cells between it and the 32).
- There's only one way to do the 7.
- The 36 can only be done as a 4x9 rectangle, and there are only two ways to place it, which force some cells.
- If R1C5 or R10C5 is shaded, that forces an unshaded rectangle in the upper left or lower left corner.
- The 6 is forced.
On puzzle 1, I count 40 ambiguous cells, and 9 solutions (3 with the 20 at R9C1-R10C10, 2 with the 20 at R7C6-R10C10, 4 with the 20 at R6C7-R10C10).
On puzzle 2, I count 2 solutions with 2 ambiguous cells.
On puzzle 3, 2 solutions with 14 ambiguous cells.
On puzzle 4, 3 solutions with 3 ambiguous cells.
On puzzle 5, 4 solutions with 8 ambiguous cells.
On puzzle 6, 2 solutions with 2 ambiguous cells. (and Sny's signature)
On puzzle 7, 18 solutions with 18 ambiguous.
On puzzle 8, 2 solutions with 2 ambiguous.
On puzzle 9, 4 solutions with 10 ambiguous.
