The solution is correct.
Look, the formula E((F b) & EGa) can be rewritten to the equivalent formula (E F b)&(E G a) which is possible since EGa is a state formula and has to hold in the first state of the path that should satisfy (F b) & EGa for the entire formula E((F b) & EGa). However, the path that satisfies (F b) & EGa, and the path that satisfies Ga in EGa may not be the same.
In the example, we have
s0 |= E(F b & E G a)
since s0s1∞ |= F b & E G a
since s0s1∞ |= F b and s0s1∞ |= E G a
since s0s1∞ |= F b and s0 |= E G a
since s0s1∞ |= F b and s0s2∞ |= G a
s0 |/= E(F b & G a)
since neither s0s1∞ |= F b & G a nor s0s2∞ |= F b & G a
since neither s0s1∞ |= F b nor s0s2∞ |= G a
Clearly, I may continue to explain why the latter is the case, but I guess that you can agree from here.