NOTE: This is just the answer for a), not for b).
a) If I draw 5 cards randomly what are the chances of getting a full house?
You took out $5$ cards, so there would be $47$ cards leftover in the deck.
There are $\binom{47}{5}=1533939$ ways to choose any $5$ cards out of a $47$-card deck.
Then I used casework to find the ways to get a full house:
Case 1: 2 is chosen three times
There are $3$ $2$'s in the modified deck, so there is only $1$ way to choose $3$ out of $3$.
Subcase 1.a: 3 is chosen 2 times
There are $3$ $3$'s in the modified deck, so there would be $\binom{3}{2}=3$ ways to get a full house this way.
Subcase 1.b: Another rank(minus the king) is chosen 2 times
There would be $4$ cards to choose from, and there are $13-3=10$ other ranks, so that means that there would be $\binom{4}{2}*10=60$ ways to get a full house this way.
In total that would be $60+3=63$ ways for Case 1.
Case 2: 3 is chosen three times
This is essentially this same thing as Case 1 so the number of ways would stay the same. $63$ ways.
Case 3: Another rank(other than 2, 3, and king) is chosen 3 times
There are $4$ cards to choose from, so that would make it $\binom{4}{3}=4$ ways to choose $3$ cards out of $4$.
Subcase 3.a: 2 is chosen 2 times
There are $3$ cards with rank $2$, so that means the number of ways to choose $2$ cards out of $3$ is $\binom{3}{2}=3$ ways for this subcase.
Subcase 3.b: 3 is chosen 2 times
Essentially the same as subcase 3.a so $3$ ways.
Subcase 3.c: Another rank(not 2, 3, king, or the rank chosen for this case) is chosen 2 times
There are $9$ ranks leftover. Each of these ranks have $4$ choices to choose from, and you are choosing $2$ out of the $4$, so the number of ways for this subcase is $\binom{4}{2}*9=54$.
In total, that would be $10*4*(3+3+54)=2400$ ways for this case.
So, the final answer is: $$\frac{63+63+2400}{1533939}=\frac{2526}{1533939}\approx 0.0016467$$