Doesn't the exam specify what it means by a "proof"? Your remark on the first proposition is probably what along the lines of what would be required of you (assuming you are working with standard semantics of predicate logic e.t.c. which from the circumstances I think I can assume you are). As is suggested in another answer, and a remark on a somewhat embarrassing error in mine, you can apply the same reasoning for the second proposition, you since you can find two situations (both satisfying $P$ and $Q$), one where it is true and one where it is false. The answer would, thus, be that not information is given (more formally you would say that the statement is neither provable nor disprovable from your assumption $P$ and $Q$).
In the third case I suspect, as another answer has already pointed out, that stating that the truth tables are equivalent is probably what they want you to do. Such a proof is not a "formal proof" (that is not to say it isn't rigorous) in the technical sense of the word though, since it is a proof on the standard semantics of propositional logic, rather than a proof within some proof system.
So in case you are familiar with any of the usual formal proof system for propositional logic you can prove the third statement by applying or-elimination, basically meaning you split the assumption of $P$ or $Q$ into two cases, one where $P$ is true, and another where $Q$ is true.
In the first case you have a contradiction (since you have now assumed $\lnot P$ and $P$) which by the principle of explosion (anything follows from a contradiction) means you can conclude $Q$.
In the second case you can trivally conclude $Q$, since you assumed it. or-elimination now states that since you could conclude $Q$ from both these cases, you can conclude $Q$ from their disjunction (their disjunction meaning the statement $P$ or $Q$).
As I already said, unless they have said something about formal proofs somewhere I doubt this is what they expect, but I figured it might be worth noting for completeness :).