As you have learned, omega-automata have acceptance conditions to decide whether an infinite run is accepting or not. In particular, safety automata have a condition that demands that the run consist only of safe states. A state is thereby safe if it is declared as such, i.e., it is part of the specification to define this.
In the exam question you mention, there is only one initial state, namely q0 and the only safe states are q1 and q2. It is pointless to discuss why that is so, it is simply given like that by this exercise, and another exercise may make another choice.
Hence, in this case, none of the runs that start in the initial state will be an accepting run, so that the automaton accepts no word. Removing all unsafe states yields an automaton without initial states, so that also that one does not accept any word. The subset construction could be applied, but again it would yield an automaton without initial states which also does not accept any word.
How is the subset construction applied? Well, as usual, we start with the set of initial states which is in this case the empty set of states {}, and this one is its only successor state for any input. It is not an initial state, however, since it does not contain an initial state. You may try this also with the corresponding teaching tool and you will see the same as written here. Try it and see.
At the end, the exam problem is therefore quite trivial and could be answered without applying any determinization procedure.
