(1) Find all prime sequences $p_{1}<p_{2}<\cdots<p_{n}$, such that
$$
\left(1+\frac{1}{p_{1}}\right)\left(1+\frac{1}{p_{2}}\right) \cdots\left(1+\frac{1}{p_{n}}\right)
$$

is a positive integer.
(2) Does there exist a positive integer $n$ and integers $a_{1}<a_{2}<\cdots<a_{n}$ greater than 1, such that
$$
\left(1+\frac{1}{a_{1}^{2}}\right)\left(1+\frac{1}{a_{2}^{2}}\right) \cdots\left(1+\frac{1}{a_{n}^{2}}\right)
$$

is a positive integer?
(1) If $n \geqslant 3$, then $p_{n}>1+p_{i}, 1 \leqslant i \leqslant n-1$. Thus, $p_{n}$ is coprime with the number $\prod_{i=1}^{n-1}\left(1+p_{i}\right)$, which implies that $p_{n} \mid p_{n}+1$, leading to $p_{n}=1$, a contradiction. When $n=1$, $1+\frac{1}{p_{1}} \notin \mathbf{N}^{*}$, so the only possibility is $n=2$.

When $n=2$, by the same analysis, we should have $p_{2} \mid p_{1}+1$, which means $p_{1}=2, p_{2}=3$. Therefore, the only sequence that satisfies the condition is $\left(p_{1}, p_{2}\right)=(2,3)$.
(2) There do not exist $a_{1}, a_{2}, \cdots, a_{n}$ that satisfy the condition. Indeed, if $1<a_{1}<a_{2}<\cdots<a_{n}$ such that $A=$ $\prod_{i=1}^{n}\left(1+\frac{1}{a_{i}^{2}}\right) \in \mathbf{N}^{*}$, we set $a_{n} \leqslant m$, then
$$
1<A \leqslant \prod_{i=2}^{m}\left(1+\frac{1}{i^{2}}\right)=\prod_{i=2}^{m} \frac{1+i^{2}}{i^{2}}<\prod_{i=2}^{m} \frac{i^{2}}{i^{2}-1}=\prod_{i=2}^{m} \frac{i^{2}}{(i-1)(i+1)}=\frac{2 m}{m+1}<2 .
$$

This contradicts $A \in \mathbf{N}^{*}$.