Your expressions for a) and b) are correct. Instead of the permutation symbol that you called $P(n,6)$, you could have used $\dbinom{n}{6}$. Doing that divides the numerator and denominator by $6!$, so it makes no difference to the answer. But it's a good thing to get used to. In the future, you will often find yourself using $\dbinom{n}{k}$, while $P(n,k)$ will be less useful.
There are various approaches to c). The most simple-minded way to count is to first count the number of choices that have all of $20$, $21$, and $22$. To count these, note that we need to choose $3$ items from the remaining $37$. This can be done in $\dbinom{37}{3}$ ways. But there are $6!$ orders in which the numbers could be chosen, so the total count (if later we will use $P(40,6)$ in the denominator) is $6!\dbinom{37}{3}$.
Now count the hands that have exactly $2$ of $20$, $21$, $22$. We can decide which $2$ in $\dbinom{3}{2}$ ways, and for each way we can select the $4$ outsiders that will join them in $\dbinom{37}{4}$ ways. Finally, multiply by $6!$ in order to scramble them in all possible orders.
But I prefer not to use $P(40,6)$ in the denominator, and instead use $\dbinom{40}{6}$. Then the numerator is $\dbinom{37}{3}+\dbinom{3}{2}\dbinom{37}{4}$.
There are other ways to count. In particular, it will after a while be important to be able to handle the method of Inclusion/Exclusion described in the answer by Ross Millikan.