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

is an integer; $\square$
(2) Does there exist $n$ distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}, n \in \mathbf{N}^{\vee}$, greater than 1, such that
$$\left(1+\frac{1}{a_{1}^{2}}\right)\left(1+\frac{1}{a_{2}^{2}}\right) \cdot \cdots \cdot\left(1+\frac{1}{a_{n}^{2}}\right)$$

is an integer?
26. (1) When $n \geqslant 3$, since $p_{n}>p_{i}+1, 1 \leqslant i \leqslant n-1$, thus $p_{n}$ is coprime with the number $\prod_{i=1}^{n-1}\left(1+p_{i}\right)$, hence it requires $p_{n} \mid\left(1+p_{n}\right)$, leading to $p_{n}=1$, which is a contradiction. When $n=1$, $1+\frac{1}{p_{1}} \notin \mathbf{N}^{*}$. When $n=2$, it should have $p_{2} \mid\left(p_{1}+1\right)$, thus $p_{2}=p_{1}+1$, which can only be $p_{1}=2, p_{2}=3$.

In summary, the required sequence is only: $\left(p_{1}, p_{2}\right)=(2,3)$.
(2) The conclusion is: it does not exist.
In fact, when $1<a_{1}<a_{2}<\cdots<a_{n}$, let $a_{n} \leqslant m$, then
$$\begin{aligned}
1<A & =\prod_{i=1}^{n}\left(1+\frac{1}{a_{i}^{2}}\right) \leqslant \prod_{i=2}^{m}\left(1+\frac{1}{i^{2}}\right)=\prod_{i=2}^{m} \frac{i^{2}+1}{i^{2}}<\prod_{i=2}^{m} \frac{i^{2}}{i^{2}-1} \\
& =\prod_{i=2}^{m} \frac{i^{2}}{(i-1)(i+1)}=\frac{2 m}{m+1}<2
\end{aligned}$$

Thus, $A \notin \mathbf{N}^{*}$.