As we have ¬F¬a = Ga, and p⊕¬p = 1, it follows that S1 is always true. S2, however, can become false, as you can see with the following traces:
a : 0101...
Ga : 0000...
XGa : 0000...
!a : 1010...
G!a : 0000...
(XGa)⊕(G!a) : 0000...
Here, you can see that the ⊕ operator (just as any other boolean operator) will be applied to the corresponding points of time of the two inputs, i.e., p⊕q denotes pointwise xor of the signal p and q.