For part (a), consider the schedule of one particular player. He has to play each of the other players once and only once, which takes 5 rounds.

For part (b), notice that since there are six players, there are 3 games per round. Therefore a total of 9 points are awarded in three rounds, and these can not be evenly split among 6 players.

Part (c) is just a matter of filling in the table.

I don't see any obvious method for (d) except to start making tables: A can play either B, C, or L. If A plays B, then C has to play K, so J plays L. If A plays C, then B has to play J, so K plays L. But if A plays L, then B can play either C or J, with the third pairing being J-K or C-K respectively.

If the third round is A-B, C-K, and J-L, then A can play either C or L in the fourth round. If A plays C, then B plays J and K has to play L. Then we can have A, K, and J win in the third round and lose in the fourth or vice-versa.

But you run into problems if the third round is A-L B-J and C-K, In the fourth round, A can play B or C. If A plays B, then C can not play anyone (he has already played L, J, and K, and A,B are already playing).

- Hollywood