Frank,
I don't think I understand your question fully. Are there just 8 players? Consider the schedule from the point of view of a single player. In each foursome there are 3 opponents, so over 4 rounds they have 4*3=12 opponents. Since there is a pool of only 7 players from which opponents can be drawn, then you can do no better than oppose 5 of the players twice and oppose the remaining 2 players once.
There is a interesting alternative as we could consider a schedule where 6 of the 7 other players are opposed exactly twice and there is no match with the remaining player. The schedule below has this property.
T1 T2 T3 T4
R1 (1 2 3 4)
R1 (5 6 7 8)
R2 (2 7 8 1)
R2 (6 3 4 5)
R3 (3 8 1 6)
R3 (7 4 5 2)
R4 (4 1 6 7)
R4 (8 5 2 3)
Player pairs that don't occur within foursomes are (1 5) (2 6) (3 7) and (4 8), so use this information to keep apart people who you think, for whatever reason, are the least compatible with each other! Alternatively make 4 teams of 2 players and then have a schedule where team-mates don't play together in the same foursome. Lastly note that all players occur once in each position T1 to T4, so each player could be given the chance to tee-off first in one of their foursomes.
Hope that helps,
Ian.
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Alternative schedule added 17th June:
If it is essential to have everyone play with everyone else at least once, the following solution is good since everyone has the same pattern of opponents. Here each player opposes one player three times, three players twice and the remaining three players once.
1 6 2 3
4 8 7 5
6 2 8 4
5 3 1 7
2 4 5 1
8 7 3 6
7 5 6 2
3 1 4 8
