Define the sequence of positive integers $\left\{a_{n}\right\}$ as
$$
a_{1}=1, a_{2}=2, a_{k+2}=2 a_{k+1}+a_{k}(k \geqslant 1) \text {. }
$$
(1) Find all positive real numbers $\beta$ such that there are infinitely many pairs of positive integers $(p, q)$ satisfying
$$
\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^{2}} ;
$$
(2) Find all positive real numbers $\beta$ such that there are only finitely many pairs of positive integers $(p, q)$ satisfying
$$
\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^{2}},
$$
and there does not exist any $n$ such that $q=a_{n}$.
First, we provide two lemmas.

Lemma 1 The positive integer solutions of the Pell equation $\left|p^{2}-2 q^{2}\right|=1$ are
$$
p=b_{n}, q=a_{n}\left(n \in \mathbf{Z}_{+}\right),
$$

where $b_{1}=1, b_{2}=3, b_{k+2}=2 b_{k+1}+b_{k}(k \geqslant 1)$.
Lemma 2 The positive integer solutions of the Pell equation $\left|p^{2}-2 q^{2}\right|=2$ are $p=2 a_{n}, q=b_{n}\left(n \in \mathbf{Z}_{+}\right)$.
Both lemmas can be proven using the basic solution method of Pell equations.
It is not difficult to calculate that
$$
\begin{array}{l}
\lim _{n \rightarrow+\infty} a_{n}^{2}\left|\frac{b_{n}}{a_{n}}-\sqrt{2}\right|=\lim _{n \rightarrow+\infty}\left|a_{n} b_{n}-\sqrt{2} a_{n}^{2}\right| \\
=\lim _{n \rightarrow+\infty} \frac{a_{n}}{b_{n}+\sqrt{2} a_{n}}=\frac{\sqrt{2}}{4} .
\end{array}
$$

Thus, we conjecture that $\beta$ in (1) is all real numbers not less than $\frac{\sqrt{2}}{4}$. In fact,
$$
\begin{array}{l}
\left|\frac{p}{q}-\sqrt{2}\right|1)
$$
$\Leftrightarrow p=2 a_{n}, q=b_{n}$ ( $n$ is an odd positive integer greater than 1).
According to the limit expression, we know that when $01$ the smallest term occurs when $n=3$. At this time,
$$
\beta=70-49 \sqrt{2} \text {. }
$$

Therefore, the range of $\beta$ is $\left(70-49 \sqrt{2}, \frac{\sqrt{2}}{2}\right)$.