Answer:
From what I understand of your problem:
1) P(X = i) where i is the number of ones amongst the five cards Jesse gets in any round.
Thus function of X => $$P(X=i) = \frac{{4\choose i}\times {(28)\choose (5-i)}}{{32\choose 5}} , i = 0,1,2,3,4$$
2) P(three cards being the same in the draw) =$$\frac{{8\choose 1}\left({4\choose 3}{28\choose 2}+{4\choose 4}{28\choose 1}\right)}{{32\choose 5}}$$
3) Part three is an application of negative binomial distribution.
Here In each round Jesse will get all cards in a round the same number $$= \frac{{8\choose 1}\left({4\choose 3}{28\choose 2}+{4\choose 4}{28\choose 1}\right)}{{32\choose 5}} = p$$ In each round Jesse will get all cards different than that in a round $$= 1 - p = q$$
Proabability that she will need two rounds of success in the six game
$${(4+2-1)\choose 4} q^4 p^2 = ({(5)\choose 4}q^4 p^2$$