Answers to Harkinian:

For the first one you are close enough. You have taken correct the number of different methods that the probability asked can happen. However the number of ways you have set for each of them is wrong. For example, in the second one, if you put 3 in A, 2 in B and 1 in C, the total ways to put 3 in A are not 6, as you write, but 10, since 5 choose 3 equals to 10.
Revise it and send me the complete answer, as all 3 of the different terms of your answer have mistaken coefficients.

For the second one, you should try another thinking process. How could the mean value be 3.30? It should be at least 6, since I'm asking for how many times you have to submit in order to get every grade at least once. For example, a possible grade outcome is {B, C+, A-, B, B+, B}, but in such outcome you have got only 4 of the 6 grades. We need the mean value of submissions that you need to make so as to get each grade at least once. It has to be -much- more than 6, so try again.

For the third one, the problem does not suggest that the proportion is equal to p. It suggests that every person has a (fixed) probability of p staying alive after many years. However, since conditioning is considered, obviously the probability of ONE person being alive now is m/2n (and NOT (2n-m)/2n), and your answer will be independent of p. However, since you consider such for one person, the next person you consider should have a probability of being alive equal to (m-1)/(2n-1). This makes the mean value harder to calculate and indicates that the calculation should be approached differently. The problem is the most challenging of the 3.