See the answer to https://q2a.cs.uni-kl.de/1051/sheet-8-exercise-1 for a general discussion about what is needed for the solution.
In your case, you have S1 = [[b SU a] WU c] and S2 = [[b SU a] SU c] and it is therefore clear that S2->S1 is valid (since SU implies WU). Hence, we have to find two nontrivial counterexamples for S1->S2, i.e., two formulas phi1 and phi2 such that the following are valid
phi1 -> [[b SU a] WU c] & ![[b SU a] SU c]
phi2 -> [[b SU a] WU c] & ![[b SU a] SU c]
Looking at the counterexample generated by the tool, you can see that phi1 := G(a&!b&!c) is such a formula. Your second formula is not correct since trying to prove
(F !(!c&a&!b)) -> [[b SU a] WU c] & ![[b SU a] SU c]
yields as a counterexample a path where a, b, and c are always false.
Find a second one that meets the above requirements. It may not necessarily be a formula starting with G or F, this is nowhere mentioned.
To give a few further hints, consider your formulas S1 = [[b SU a] WU c] and S2 = [[b SU a] SU c]. If eventually c holds, both formulas are equivalent, i.e.,
(F c) -> ([[b SU a] WU c] <-> [[b SU a] SU c])
is valid. So, all counterexamples must imply G!c. However, that is not enough since if just G!c would hold, then both formulas S1 and S2 are simply false. Now note that if c is never true, [[b SU a] WU c] will become equivalent to G[b SU a] which implies that infinitely often "a" must hold. From here, you may guess another candidate formula.
