The following is a semantic (model-theoretic) argument for the quantifiers question. (I cannot provide a syntactic argument, since I do not know what axioms, rules of inference are used in your course. There is wide variability.)
So we use the following criterion: A sentence in a certain language $L$ is valid if it is true in every $L$-structure $M$. Note that an $L$-structure has, by definition, a non-empty underlying set.
We will assume that $P$ has only one free occurrence of a variable. If in an $L$-structure $M$ for the appropriate language $L$, there really is a $b$ such that $P(b)$ is true in $M$, then $\exists y\forall x(P(x)\to P(y))$ is true in $M$, since $\forall x(P(x)\to P(b))$ is true in $M$. Thus the sentence you asked about is true in $M$. The antecedent is irrelevant.
If there is no $n$ in $M$ such that $P(n)$ is true, then again $\exists y\forall x(P(x)\to P(y))$ is true in $M$. For let $b$ be some fixed element of $M$.
Note that $P(b)$ is false in $M$. But for any $m$ in $M$, $P(m)$ is false in $M$. It follows that for any $m$, $P(m)\to P(b)$ is true in $M$. So again the sentence you asked about is true in $M$, and again the antecedent is irrelevant.
Comment: The whole thing hinges on the fact that $X\to Y$ is true whenever $Y$ is true, and also whenever $X$ is false. The connective "implies" in ordinary English is not simply truth-functional. That is a frequent source of confusion.
Added: For the first part, we will use a variant of one of the approaches you tried. You tried to show that if the left side is true, then the right side is true. That can be done. But to me it is somewhat easier to show (in this case) that if the right side of the implication is false, then the left side is false.
We want to show that
$$((p \lor q) \land (\lnot p \lor r)) \to (q \lor r)$$
is tautologous. Think of your sentence as an implication $A \to B$. For this to be false, we need $q\lor r$ false and $(p \lor q) \land (\lnot p \lor r)$ true. For $q\lor r$ to be false, we need $q$ false and $r$ false. So suppose $q$ and $r$ are false.
The only way then for $p\lor q$ to be true is if $p$ is true. The only way for $\lnot p \lor q$ to be true is if $p$ is false. So for both of them to be true, we need $p$ to be simultaneously true and false! We conclude that if $p$ and $q$ are false, then $(p \lor q) \land (\lnot p \lor r)$ is false, so the implication is always true.