I don't believe that this is possible. If you meet the criteria regarding the pros (don't play together, and play 10 members each), then you would have to have pairs of members who play each other twice. I think there would be as many as 8 of these pairs. The best solutions will have some asymmetry regarding the pros and the distribution of players over the threesomes and foursomes. For example consider the following:
( 3 12 6) ( 9 8 11) (4 13 1) (5 2 7 10)
(10 1 6) ( 9 12 7) (4 8 2) (5 13 3 11)
(11 2 6) (13 9 10) (4 7 3) (5 8 1 12)
( 2 1 3) ( 7 13 8) (5 9 6) (4 10 11 12)
Where players 4 and 5 are the two pros. Pro 4 meets 9 different members, while Pro 5 meets all 11 members. No pair of players meeet twice. The members have different numbers of opponents too, 10, 11 and 12 have two foursomes, while 6 and 9 are never in a foursome. Hope that helps.
Ian