It is better to ask one question per post. Here is the answer to Q1. Assume that the pair $(x,y)$ satisfies $(1)$. Then $(x,ay)$ is a solution of $X^2-aY^2=1$. Let us restrict to $a=u^2-1$, where $u>1$ is an integer. Then the positive integer solutions of $X^2-aY^2=1$ are the pairs $(X_n,Y_n)$ satisfying $$X_n+Y_n\sqrt{a}=(u+\sqrt{a})^n,\qquad n\in\mathbb{Z}_{\geq 1}.$$ Hence $(x,ay)=(X_n,Y_n)$ for some positive integer $n$, which means that $$ay=Y_n=\sum_{\substack{0\leq k\leq n\\\text{$k$ odd}}}\binom{n}{k}u^{n-k}a^{(k-1)/2}.$$ On the right-hand side each term with $k>1$ is divisible by $a$, hence the term with $k=1$ is also divisible by $a$. That is, $a\mid nu^{n-1}$. So $a\mid n$, and then $$x=X_n>\frac{X_n+Y_n\sqrt{a}}{2}=\frac{(u+\sqrt{a})^n}{2}>\frac{(2\sqrt{a})^a}{2}>a^{a/2}.$$ So $x$ is not $a^{O(1)}$ when $a=u^2-1$.

It is better to ask one question per post. Here is the answer to Q1. 

Assume that $(x,y)$ is a positive integer solution of $(1)$. Then $(x,ay)$ is a solution of $X^2-aY^2=1$. Let us restrict to $a=u^2-1$, where $u>1$ is an integer. Then the positive integer solutions of $X^2-aY^2=1$ are the pairs $(X_n,Y_n)$ satisfying $$X_n+Y_n\sqrt{a}=(u+\sqrt{a})^n,\qquad n\in\mathbb{Z}_{\geq 1}.$$ Hence $(x,ay)=(X_n,Y_n)$ for some positive integer $n$, which means that $$ay=Y_n=\sum_{\substack{0\leq k\leq n\\\text{$k$ odd}}}\binom{n}{k}u^{n-k}a^{(k-1)/2}.$$ On the right-hand side each term with $k>1$ is divisible by $a$, hence the term with $k=1$ is also divisible by $a$. That is, $a\mid nu^{n-1}$. So $a\mid n$, and then $$x=X_n>\frac{X_n+Y_n\sqrt{a}}{2}=\frac{(u+\sqrt{a})^n}{2}>\frac{(2\sqrt{a})^a}{2}>a^{a/2}.$$ So $x$ is not $a^{O(1)}$ when $a=u^2-1$.

It is better to ask one question per post. Here is the answer to Q1. Assume that the pair $(x,y)$ satisfies the Pell equation in question. Then $(x,ay)$ is a solution of $X^2-aY^2=1$. Let us restrict to $a=u^2-1$, where $u>1$ is an integer. Then the positive integer solutions of $X^2-aY^2=1$ are the pairs $(X_n,Y_n)$ satisfying $$X_n+Y_n\sqrt{a}=(u+\sqrt{a})^n,\qquad n\in\mathbb{Z}_{\geq 1}.$$ Hence $(x,ay)=(X_n,Y_n)$ for some positive integer $n$, which means that $$ay=Y_n=\sum_{\substack{0\leq k\leq n\\\text{$k$ odd}}}\binom{n}{k}u^{n-k}a^{(k-1)/2}.$$ On the right-hand side each term with $k>1$ is divisible by $a$, hence the term with $k=1$ is also divisible by $a$. That is, $a\mid nu^{n-1}$. So $a\mid n$, and then $$x=X_n>\frac{X_n+Y_n\sqrt{a}}{2}=\frac{(u+\sqrt{a})^n}{2}>\frac{(2\sqrt{a})^a}{2}>a^{a/2}.$$ So $x$ is not $a^{O(1)}$ when $a=u^2-1$.

It is better to ask one question per post. Here is the answer to Q1. Assume that the pair $(x,y)$ satisfies $(1)$. Then $(x,ay)$ is a solution of $X^2-aY^2=1$. Let us restrict to $a=u^2-1$, where $u>1$ is an integer. Then the positive integer solutions of $X^2-aY^2=1$ are the pairs $(X_n,Y_n)$ satisfying $$X_n+Y_n\sqrt{a}=(u+\sqrt{a})^n,\qquad n\in\mathbb{Z}_{\geq 1}.$$ Hence $(x,ay)=(X_n,Y_n)$ for some positive integer $n$, which means that $$ay=Y_n=\sum_{\substack{0\leq k\leq n\\\text{$k$ odd}}}\binom{n}{k}u^{n-k}a^{(k-1)/2}.$$ On the right-hand side each term with $k>1$ is divisible by $a$, hence the term with $k=1$ is also divisible by $a$. That is, $a\mid nu^{n-1}$. So $a\mid n$, and then $$x=X_n>\frac{X_n+Y_n\sqrt{a}}{2}=\frac{(u+\sqrt{a})^n}{2}>\frac{(2\sqrt{a})^a}{2}>a^{a/2}.$$ So $x$ is not $a^{O(1)}$ when $a=u^2-1$.