As stated the problem can not be solved since the total number of byes is 16*4=64, a number that is not divisible by 12, so some players will have more byes (and fewer games) than others. The best compromise would be to play 15 weeks. I can find a schedule for this where no partnerships are repeated and where pairs oppose no more than twice.
Court 1 Court 2 Byes
( 1 8 v 7 5) ( 4 6 v 9 11) (3 12 2 10)
( 2 9 v 8 6) ( 5 4 v 7 12) (1 10 3 11)
( 3 7 v 9 4) ( 6 5 v 8 10) (2 11 1 12)
(11 10 v 3 12) ( 1 7 v 2 6) (4 9 5 8)
(12 11 v 1 10) ( 2 8 v 3 4) (5 7 6 9)
(10 12 v 2 11) ( 3 9 v 1 5) (6 8 4 7)
(12 1 v 3 8) (11 5 v 2 4) (9 7 6 10)
(10 2 v 1 9) (12 6 v 3 5) (7 8 4 11)
(11 3 v 2 7) (10 4 v 1 6) (8 9 5 12)
( 9 6 v 5 12) (10 3 v 7 8) (1 4 2 11)
( 7 4 v 6 10) (11 1 v 8 9) (2 5 3 12)
( 8 5 v 4 11) (12 2 v 9 7) (3 6 1 10)
( 1 3 v 11 6) (10 9 v 8 4) (2 5 12 7)
( 2 1 v 12 4) (11 7 v 9 5) (3 6 10 8)
( 3 2 v 10 5) (12 8 v 7 6) (1 4 11 9)