Let's answer the question 2:
We can write a generating function for the number of combinations. The number $1$ is present either with a $+$ or a $-$, which gives 1 way of making $1$ and 1 way of making $-1$. We will encode these data with a polynomial $z + z^{-1}$. Similarly, we writer $z^2 + z^{-2}$ for the number 2, etc. up to $z^7 + z^{-7}$. The product of all of those polynomials, $G = \prod_{n=1}^7 (z^n + z^{-n})$, is also a polynomial, and the coefficient at any $z^{k}$ is exactly the number of ways to obtain the sum of $k$. (If the term with $z^k$ is not present at all, the coefficient is zero and the sum of $k$ cannot be achieved at all.)
We can immediately answer the question 1 if we notice that
$G = \prod_{n=1}^7 (z^n + z^{-n})$ can be written as $G = z^{-28} \prod_{n=1}^7 (z^{2n}+1)$. This contains all even powers between $z^{-28}$ and $z^{28}$ (the coefficients are always only a sum of some +1's, i. e. nonzero); so all even numbers can be achieved and no odd number can be achieved.
If there were more than 7 numbers, then it would make sense to try to come up with some algebraic tricks, but
here we can just put it into Sage to get $$G = z^{28} + z^{26} + z^{24} + 2 \, z^{22} + 2 \, z^{20} + 3 \, z^{18} + 4 \, z^{16} + 5 \, z^{14} + 5 \, z^{12} + 6 \, z^{10} + 7 \, z^{8} + 7 \, z^{6} + 8 \, z^{4} + 8 \, z^{2} + 8 + \frac{8}{z^{2}} + \frac{8}{z^{4}} + \frac{7}{z^{6}} + \frac{7}{z^{8}} + \frac{6}{z^{10}} + \frac{5}{z^{12}} + \frac{5}{z^{14}} + \frac{4}{z^{16}} + \frac{3}{z^{18}} + \frac{2}{z^{20}} + \frac{2}{z^{22}} + \frac{1}{z^{24}} + \frac{1}{z^{26}} + \frac{1}{z^{28}}.$$ Other people have already explained why there can be only even sums. This shows that, out of those, the sums ±28 to ±24 have a single combination of signs, ±22 and ±20 two, ±18 three, ±16 four, ±14 and ±12 five, ±10 six, ±8 and ±6 seven and ±4, ±2 and 0 eight.
