A few comments about the second problem...
1. I agree with the 9.9% answer (540/1716)^2
2. Pairing the matches is interesting. I originally read the second part of the problem to mean that. for example, Team A's Males are assigned to a single group and no other Male team is transferred intact AND Team A's Females are assigned to a single group and no other Female team's members are reassigned to the same group. That is, a full original team is transferred intact while all others are split.
Let
M6 = Male group of 6 members
M7 = Male group of 7 members
F6 = Female group of 6 members
F7 = Female group of 7 members
Individually, we have the following
Team of 6 Team of 7
A 27 108
B 27 108
C 27 108
D 27 108
Total 108 432
as noted by others. A team is 4 times more likely to transfer intact to the group of 7 than the group of 6.
There are 291,600 (540^2) total intact teams, broken down as
M6F6 = 729 x 4 = 2916
M6F7 = 2916 x 4 = 11664
M7F6 = 2916 x 4 = 11664
M7F7 = 11664 x 4 = 46656
Total matched pairs (only a single original team have both M anf F members transfer intact). = 72900
Total non-matched pairs (only one team each transfers, but not from the same original grouping, e.g. A males and C females) = 218700
M6F6 = 2187 x 4 = 8748
M6F7 = 8748 x 4 = 34992
M7F6 = 8748 x 4 = 34992
M7F7 = 34992 x 4 = 139968
Following probability rules of independent events, note that combined M and F transfers are 16x more likely to transfer to the team of 7 rather than to the team of 6.
Looking at the combinations, it is easy to see, but my first intuition would have been a much smaller difference. Perhaps someone can devise a parlor trick to take advantage of this fact.
