Yes, you'll need to use 10_C_3, because all of the chances are the amount of combinations that meet the condition in the question, divided by the total amount of combinations.

Now you know this I'll give you some hints to work it out:

a) A has to be in the committee, in how many ways can the leftover 2 members be chosen?

b) Does this one differ from A (in a probabilistical way)?

c) A and B are in there, and one other, in how many ways can that person be chosen?

d) This depends a bit on how the question is, is it one of them only, or can it also be both of them. If it's the second one, this is the formula to get you started.

This is meant to help you on your way, if you still don't understand it, don't hesitate to ask help.