1.a. 5^5 = 31125

1.b. 5! = 60

2. There 6*7 - 42 different ways to sell 1 math and 1 science book, 6*4 = 24 ways to sell 1 math and 1 economics book, and 7*4 = 28 ways to sell 1 science and 1 economics book, for a total of 94.

3. Lets call the 2 people who will not serve together John and Mary. There are (8 choose 3) * (6 choose 3) = 1120 total committees, and (7 choose 2) * (5 choose 2) = 210 that that have both John and Mary, so there are 1120 - 210 = 910 possible committees that do not include both John and Mary. Or you could also calculate the number of committees that Mary is on but not John (7 choose 2) * (5 choose 3) = 210, plus the number of committees that John is on but not Mary (5 choose 2) * (7 choose 3) = 350, plus the number of committees that both Mary and John are not on (7 choose 3) * (5 choose 3) = 350, for a total of 910.

4. Lets call the 2 friends who will only attend together John and Mary. There are (6 choose 5) = 6 ways to invite 5 friends that would not include John or Mary, and there are (6 choose 3) = 20 ways to choose 5 friends that would include both John and Mary, so there are 26 choices for invites to the party. Or you could approach by calculating the total number of invites possible (8 choose 5) = 56 and subtracting the number that would include only John (6 choose 4) = 10 and the number that would include only Mary (6 choose 4) = 10. So the total would e 56 - 10 -10 = 26.

I am trying to get better at these types of problems myself, so above is what I think are correct answers but not certain myself.