The hardest problem to solve is 12 players for 8 rounds and I think the following is about the best that can be done:
(12 1 v 8 5) (11 3 v 7 9) ( 6 4 v 2 10)
( 9 10 v 11 8) ( 3 6 v 5 1) (12 7 v 2 4)
( 3 2 v 5 9) (10 12 v 1 7) ( 6 11 v 8 4)
( 5 3 v 10 4) ( 9 11 v 12 2) ( 7 6 v 1 8)
( 7 8 v 5 2) ( 4 3 v 1 11) ( 9 12 v 6 10)
( 1 6 v 2 9) ( 4 12 v 3 8) (11 7 v 5 10)
(10 11 v 1 2) ( 8 9 v 4 5) ( 6 12 v 7 3)
( 2 8 v 3 10) (11 12 v 6 5) ( 9 1 v 4 7)
All the partnerships are different and pairs play in opposition either once or twice.
Some of the other problems that you have listed are close to a classic whist design, so the best solution is to start with one of those and modify it as necessary.
You can find the whist schedules by clicking on the 'schedules' button above and then choosing 'whist'.
For example
8 players and 8/10 weeks: add rounds to the fully balanced 7 round whist schedule for 8 players.
11 players leaves 3 byes per round, and you would need 11 rounds to give everyone the same number of games (8 games + 3 byes). Is it possible to play with one table of 3 - in which case delete player 12 from the 12 player whist?