Consider an accepting run through a co-Büchi automaton: Since it must satisfy the acceptance condition FGφ, it must finally end with a loop in φ-states (which may not just be a self-loop). If the φ-states are a strongly connected component (i.e., non-φ-states may reach them, but you cannot go back from the φ-states to the non-φ-states) then FGφ and Fφ means the same.
Otherwise, the trick is as you say and how it is described in the solution of the mentioned exam question: Make a copy of the φ-states that inherit all transitions from non-φ-states towards them, but not those from the new φ-states to the non-φ-states. This makes the automaton nondeterministic, but now you can say that the newly added states should be the accepting states. Why that? The automaton has no new runs, but now the φ-states form a connected component, so that you can replace now FG by F.