DetFG is more expressive than DetF, you mean that DetFG is equally expressive as NDetF. For this reason, every NDetF automaton can be converted to a DetFG automaton. However, it is not the case that you can just apply the breakpoint construction to the given NDetF automaton. You first have to convert it to an equivalent NDetFG automaton, and then you can apply the breakpoint construction.

In part a) of the exam problem, it was learned that we can make all states accepting for the coBüchi condition. Then, you can also make it a safety automaton where all four states are safe states. That one can be determined by the subset construction, so that this example can even be given as DetG automaton.

For particular variants of the constructions, see the updated solutions of the exam paper, in particular, you cannot apply the breakpoint construction to the generate a DetFG, since the word where a is always false will no longer be accepted.