(1) Find all natural numbers $n$ such that $S_{n}=9+17+25+\cdots+(8 n+1)$ is a perfect square;
(2) Find all natural numbers $n$ such that $T_{n}=5+11+17+\cdots+(6 n-1)$ is a perfect square.
(1) Since $S_{n}=9+17+25+\cdots+(8 n+1)=n(4 n+5)$, and $(n, 4 n+5)=(n, 5)$. If $(n, 5)=1$, and $S_{n}$ is a perfect square, then $n, 4 n+5$ are both perfect squares. Let's assume $n=k^{2}$, $4 n+5=l^{2}$, thus
$4 n+5=4 k^{2}+5=l^{2}$, which means $5=(l-2 k)(l+2 k)$, so $l-2 k=1, l+2 k=5$, giving $l=3, k=1$. At this point, $n=1$.

If $(n, 5)=5$, let $n=5 k$, then $S_{n}=n(4 n+5)=5 k(20 k+5)=5^{2} k(4 k+1)$. Since $(k, 4 k+1)=1$, if $S_{n}$ is a perfect square, then $k, 4 k+1$ are both perfect squares, but if $k$ is a perfect square, $4 k$ is a perfect square, and $4 k+1$ is not a perfect square, so in this case $S_{n}$ is not a perfect square.
(2) Since $T_{n}=5+11+17+\cdots+(6 n-1)=n(3 n+2)$, and $(n, 3 n+2)=(n, 2)$. If $(n, 2)=1, T_{n}$ is a perfect square, then $n, 3 n+2$ are perfect squares, let's assume $n=k^{2}, 3 n+2=l^{2}$, we get $3 k^{2}+2=l^{2}$, which means $l^{2} \equiv 2(\bmod 3)$. Note that, a perfect square divided by 3 has a remainder of 0 or 1. Therefore, $3 k^{2}+2=l^{2}$ has no integer solutions. At this point, $T_{n}$ cannot be a perfect square.
If $(n, 2)=2$, let $n=2 k$, then $T_{n}=4 k(3 k+1)$. Since $(k, 3 k+1)=1$. If $T_{n}$ is a perfect square, then $k, 3 k+1$ are perfect squares. Let $k=m^{2}, 3 k+1=l^{2}$, we get $l^{2}-3 m^{2}=1$. The smallest positive integer solution to this Pell's equation is $l_{1}=2, m_{1}=1$. Therefore, all positive integer solutions are
$$
\begin{array}{c}
l_{h}=\frac{1}{2}\left(\left(l_{1}+\sqrt{3} m_{1}\right)^{h}+\left(l_{1}-\sqrt{3} m_{1}\right)^{h}\right), \\
m_{h}=\frac{1}{2 \sqrt{3}}\left(\left(l_{1}+\sqrt{3} m_{1}\right)^{h}-\left(l_{1}-\sqrt{3} m_{1}\right)^{h}\right), h \in \mathbf{N} .
\end{array}
$$

Thus, for all $n=2 k=2 m_{h}^{2}=\frac{1}{6}\left(\left(l_{1}+\sqrt{3} m_{1}\right)^{h}-\left(l_{1}-\sqrt{3} m_{1}\right)^{h}\right)^{2}=\frac{1}{6}((2+$ $\left.\sqrt{3})^{h}-(2-\sqrt{3})^{h}\right)^{2}, h=1,2,3, \cdots, T_{n}$ is a perfect square.