The solution given for the exam ist correct, and yours is unfortunately wrong. To compute the transitions of state ({s2,s3},{s3}), we have to consider the transitions from states s2 and s3 which are as follows: s2--*-->s2, s2--*-->s3 and s3--!b-->s3. Hence, with input !b (regardless of a), we have ({s2,s3},{s3}) --!b--> ({s2,s3},{s3}) and with input b (again regardless of a), we have ({s2,s3},{s3}) --b--> ({s2,s3},{}) since we have to compute the existential successors in both components, and have to intersect the second component with {s3}.
It is easily seen that your solution cannot be correct, since your automaton is not deterministic. Note that a and b are boolean variables that symbolically encode the four inputs !a&!b, !a&b, a&!b, and a&b. Having outgoing transitions for a as well as for condition b (and !b) gives us choices for the input a&b: Since a holds, we could remain in state ({s2,s3},{s3}) of your automaton, and since b holds, we may switch to state ({s2,s3},{}). Hence, your automaton is not deterministic, and can therefore be not a correct result of a determinization procedure.
Accepting states of the automaton are those where the second component is not empty. That is only the case for state ({s2,s3},{s3}) and in particular not for state ({s2,s3},{}).
I think that you mean a sink state instead of a sync state. But why should we have a sink state? It can be generated by the determinization procedure if needed, but it is not generated here, since it is not needed here.
