your ideas for part a) and c) are correct. for part b) let if all are > 0, then and is a partition of n. if at least one of is 0, say then

and is a partition of 2n - 1, and you're done. this also tells us how to find partitions of 2n when the partitions of n and 2n - 1 are given. for example for n = 3:

partitions of 3: 1 + 1 + 1, 1 + 2.

partitions of 5: 1 + 2 + 2, 1 + 4, 1 + 1 + 1 + 2, 1 + 1 + 1 + 1 + 1.

now to find the partitions of 2n = 6, we multilpy every partition of n = 3 by 2, and add 1 unit to every partition of 2n - 1 = 5 to get all 6 partitions of 6:

2 + 2 + 2, 2 + 4, 1 + 1 + 2 + 2, 1 + 1 + 4, 1 + 1 + 1 + 1 + 2, 1 + 1 + 1 + 1 + 1 + 1.