Five. (20 points) Given $f(x)=\frac{1}{4^{x}+2}(x \in \mathbf{R})$, $P_{1}\left(x_{1}, y_{1}\right)$ and $P_{2}\left(x_{2}, y_{2}\right)$ are two points on the graph of the function $y=f(x)$, and the x-coordinate of the midpoint $P$ of the line segment $P_{1} P_{2}$ is $\frac{1}{2}$.
(1) Prove that the y-coordinate of point $P$ is a constant;
(2) If the general term formula of the sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=f\left(\frac{n}{m}\right) \quad\left(m \in \mathbf{N}_{+}, n=1,2, \cdots, m\right),
$$

find the sum of the first $m$ terms $S_{m}$ of the sequence $\left\{a_{n}\right\}$;
(3) If the inequality $\frac{a^{m}}{S_{m}}<\frac{a^{m+1}}{S_{m+1}}$ holds for all $m \in \mathbf{N}_{+}$, find the range of the real number $a$.
(1) From $\frac{x_{1}+x_{2}}{2}=\frac{1}{2}$, we know $x_{1}+x_{2}=1$. Then
$$
\begin{array}{l}
y_{1}+y_{2}=\frac{1}{4^{x_{1}}+2}+\frac{1}{4^{x_{2}}+2} \\
=\frac{\left(4^{x_{2}}+2\right)+\left(4^{x_{1}}+2\right)}{4^{x_{1}+x_{2}}+2 \times 4^{x_{1}}+2 \times 4^{x_{2}}+4} \\
=\frac{4^{x_{1}}+4^{x_{2}}+4}{2\left(4^{x_{1}}+4^{x_{2}}+4\right)}=\frac{1}{2} .
\end{array}
$$

Therefore, the y-coordinate of point $P$ $\frac{y_{1}+y_{2}}{2}=\frac{1}{4}$ is a constant.
$$
\begin{array}{l}
\text { (2) Given } S_{m}=a_{1}+a_{2}+\cdots+a_{m} \\
=f\left(\frac{1}{m}\right)+f\left(\frac{2}{m}\right)+\cdots+f\left(\frac{m-1}{m}\right)+f(1), \\
\text { and } S_{m}=a_{m-1}+a_{m-2}+\cdots+a_{1}+a_{m} \\
=f\left(\frac{m-1}{m}\right)+f\left(\frac{m-2}{m}\right)+\cdots+f\left(\frac{1}{m}\right)+f(1),
\end{array}
$$

Adding the two equations, we get
$$
\begin{aligned}
2 S_{m}= & {\left[f\left(\frac{1}{m}\right)+f\left(\frac{m-1}{m}\right)\right]+\left[f\left(\frac{2}{m}\right)+f\left(\frac{m-2}{m}\right)\right]+} \\
& \cdots+\left[f\left(\frac{m-1}{m}\right)+f\left(\frac{1}{m}\right)\right]+2 f(1) .
\end{aligned}
$$

Since $\frac{k}{m}+\frac{m-k}{m}=1(k=1,2, \cdots, m-1)$, we have $f\left(\frac{k}{m}\right)+f\left(\frac{m-k}{m}\right)=\frac{1}{2}$,

and $f(1)=\frac{1}{6}$, thus, $S_{m}=\frac{1}{12}(3 m-1)$.
(3) From $\frac{a^{m}}{S_{m}}>0$, we get $a^{m}>0$. When $a^{m}>0$, since $a^{m}>0$, then from equation (1) we get
$$
a>\frac{3 m+2}{3 m-1}=1+\frac{3}{3 m-1} \text {. }
$$

Since $\frac{3}{3 m-1}$ decreases as $m$ increases, when $m=1$, $1+\frac{3}{3 m-1}$ reaches its maximum value $\frac{5}{2}$. Therefore, $a>\frac{5}{2}$.