for 1) (assuming coin is 50/50), use P(H) = C(9,H) * (0.5)^H * (0.5)^(9-H), H is the number of heads

2a)there are C(10,6) ways to choose 6 students from the group of 10, not considering order (this is basically just the definition of the choose function)

b) now 2 of the six slots are filled, and again not considering order there is 1 way to do this. now there is 8 remaining students, so the total number of ways is 1x C(8,4) = C(8,4)

c) now you're just picking from a group of 9 students, so there are C(9,6) ways

d) i'm assuming this question means that when you pick the 6 students, you can't pick both A and B. there are two cases: one where you pick 1 of A and B, and one where you pick neither. for the first case, you have 2 options to pick the first student (A or B), and then C(8,5) ways to pick the other five studnets, giving 2xC(8,5) ways

for the second case, you have only 8 students to pick from, so there are C(8,6) ways

so in total, there are 2xC(8,5) + C(8,6) ways

hopefully thats right