Yes, this problem is probably small enough to find a solution by hand. Running it through a search program however, gives me the following:
T1 T2
(A E) (- -)
(F G) (B C)
(C E) (A D)
(D F) (B G)
(A C) (E G)
(B F) (C D)
(B E) (A G)
(D G) (C F)
(A B) (D E)
(C G) (E F)
(B D) (A F)
Team A has the best timings, always having a rest. Teams B, E and F have back-to-back play once, teams C, D and G, have two occurences of back-to-back play.
I believe that teams play 3 or 4 times at time 1 (T1) and 2 or 3 times at time 2 (T2).