What I am thinking is for part "a" to do (4/52)*(3/51) since the card is not being replaced. But the word "exactly" is throwing me off. Am I on the right track?
Pick a card; keep it; pick another card. Find the probability of getting
a. exactly 2 queens.______
b. exactly one queen.______
c. no queens.________
First, one thing to note is that the question indicates no replacement this is usually a good hint for you to think about the hypergeometric distribution as that is the only probability distribution with no replacement.
In general and as indicated by Wikipedia, the distribution takes the following form:
$\frac{{K \choose k} {N-K \choose n-k}}{N \choose n} $
Where:
$N$ is the population size,
$K$ is the number of success states in the population,
$n$ is the number of draws (i.e. quantity drawn in each trial),
$k$ is the number of observed successes
Applying this formula for your problem we get the following general formula: $\frac{{4 \choose k} {48 \choose 2-k}}{52 \choose 2}$ for $k=0,1,2$
$N$ and $n$ are fixed since we are always talking $2$ cards out of $52$ and $K$ is fixed since in any standard deck of cards there is always $4$ queens.
Now going back to your main question and applying the formula, we get the following:
1.$\frac{{4 \choose 2} {48 \choose 0}}{52 \choose 2}$ since both our cards are queen we are not picking anything from the other $48$.
2.$\frac{{4 \choose 1} {48 \choose 1}}{52 \choose 2}$ since we are picking one queen out of the $4$ we have and $1$ other card out of the $48$ we have.
3.$\frac{{4 \choose 0} {48 \choose 2}}{52 \choose 2}$ since we are picking no queens and both our cards are from the $4$8 we have that aren't queens.