Example 9 Given the function $f(x)=\frac{1}{4^{x}+2}(x \in \mathbf{R})$, points $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right)$ are two points on the graph of the function $f(x)$, and the x-coordinate of the midpoint $P$ of the line segment $P_{1} P_{2}$ is $\frac{1}{2}$.
(1) If the general term formula of the sequence $\left\{a_{n}\right\}$ is $a_{n}=f\left(\frac{n}{m}\right)(m \in \mathbf{N}, n=1,2, \cdots, m)$, find the sum of the first $m$ terms of the sequence $\left\{a_{n}\right\}$, $S_{m}$;
(2) If the inequality $\frac{a^{m}}{S_{m}}<\frac{a^{m+1}}{S_{m+1}}$ always holds when $m \in \mathbf{N}^{\vee}$, find the range of the real number $a$.
(1) From the problem, we know: $x_{1}+x_{2}=2 \times \frac{1}{2}=1$, so,
$$
\begin{aligned}
y_{1}+y_{2} & =f\left(x_{1}\right)+f\left(x_{2}\right)=\frac{1}{4^{x_{1}}+2}+\frac{1}{4^{x_{2}}+2}=\frac{4^{x_{1}}+4^{x_{2}}+4}{\left(4^{x_{1}}+2\right)\left(4^{x_{2}}+2\right)} \\
& =\frac{4^{x_{1}}+4^{x_{2}}+4}{4^{x_{1}+x_{2}}+2\left(4^{x_{1}}+4^{x_{2}}\right)+4}=\frac{4^{x_{1}}+4^{x_{2}}+4}{2\left(4^{x_{1}}+4^{x_{2}}+4\right)}=\frac{1}{2}
\end{aligned}
$$

The y-coordinate of point $P$ is $y_{P}=\frac{y_{1}+y_{2}}{2}=\frac{1}{4}$, which is a constant. Therefore,
for any natural numbers $m, n, f\left(\frac{n}{m}\right)+f\left(\frac{m-n}{m}\right)=\frac{1}{2}$ always holds.
Since $S_{m}=f\left(\frac{1}{m}\right)+f\left(\frac{2}{m}\right)+\cdots+f\left(\frac{m-2}{m}\right)+f\left(\frac{m-1}{m}\right)+f\left(\frac{m}{m}\right)$, we can consider using the method of adding in reverse order. That is, since:
$$
\begin{aligned}
S_{m} & =f\left(\frac{1}{m}\right)+f\left(\frac{2}{m}\right)+\cdots+f\left(\frac{m-2}{m}\right)+f\left(\frac{m-1}{m}\right)+f\left(\frac{m}{m}\right) \\
& =f\left(\frac{m}{m}\right)+f\left(\frac{m-1}{m}\right)+f\left(\frac{m-2}{m}\right)+\cdots+f\left(\frac{2}{m}\right)+f\left(\frac{1}{m}\right)
\end{aligned}
$$

Therefore, $2 S_{m}=\left[f\left(\frac{1}{m}\right)+f\left(\frac{m-1}{m}\right)\right]+\left[f\left(\frac{2}{m}\right)+f\left(\frac{m-2}{m}\right)\right]+\cdots+\left[f\left(\frac{m-1}{m}\right)+\right.$ $\left.f\left(\frac{1}{m}\right)\right]+2 f\left(\frac{m}{m}\right)=\frac{1}{2}(m-1)+2 f(1)=\frac{1}{6}(3 m-1)$, so, $S_{m}=\frac{1}{12}(3 m-1)$.
(2) From $S_{m}=\frac{1}{12}(3 m-1)$, we get $S_{m-1}=\frac{1}{12}(3 m+2)$,
Thus, $\frac{a^{m}}{S_{m}}>0$, it is only necessary that $a^{m}>0$ when $m \in \mathbf{N}$, so, it is necessary and sufficient that $\frac{1}{3 m-1}-\frac{a}{3 m+2}<\frac{3 m+2}{3 m-1}$ for $m \in \mathbf{N}$.
Let $g(m)=\frac{3 m+2}{3 m-1}\left(m \in \mathbf{N}^{*}\right)$.
Since $g(m+1)-g(m)=\frac{3 m+5}{3 m+2}-\frac{3 m+2}{3 m-1}=\frac{-9}{(3 m+2)(3 m-1)}<0$, $g(m)$ is a decreasing function, and $g(1)=\frac{5}{2}$.