Five. (20 points) A function $f(x)$, if for any triangle, as long as its three side lengths $a, b, c$ are within the domain of $f(x)$, then $f(a), f(b), f(c)$ are also the side lengths of some triangle, is called a “triangle-preserving function”.
(1) If $g(x)$ is a periodic function defined on $\mathbf{R}$ with a range of $(0,+\infty)$, prove: $g(x)$ is not a triangle-preserving function;
(2) If $h(x)=\sin x(x \in(0, M))$ is a triangle-preserving function, find the maximum value of $M$.
(1) Let $T>0$ be a period of the function $g(x)$. Since its range is $(0,+\infty)$, there exist $m > n > 0$ such that $g(m)=2, g(n)=1$.

Take a positive integer $\lambda > \frac{m-n}{T}$, then $m, \lambda T+n, \lambda T+n$ can be the three sides of a triangle. However, $g(m)=2, g(\lambda T+n)=1, g(\lambda T+n)=1$ cannot be the three sides of any triangle. Therefore, $g(x)$ is not a triangle-preserving function.
(2) The maximum value of $M$ is $\frac{5 \pi}{6}$.
On one hand, if $M > \frac{5 \pi}{6}$, we will prove that $h(x)=\sin x$ $(x \in (0, M))$ is not a triangle-preserving function.

Take $\frac{\pi}{2}, \frac{5 \pi}{6}, \frac{5 \pi}{6} \in (0, M)$. Clearly, these three numbers can be the three sides of a triangle. However, $\sin \frac{\pi}{2}=1$, $\sin \frac{5 \pi}{6}=\frac{1}{2}$, $\sin \frac{5 \pi}{6}=\frac{1}{2}$ cannot be the three sides of any triangle. Therefore, $h(x)=\sin x (x \in (0, M))$ is not a triangle-preserving function.

On the other hand, we will prove that when $M=\frac{5 \pi}{6}$, $h(x)=\sin x (x \in (0, M))$ is a triangle-preserving function.

For any triangle with sides $a, b, c$, if $a, b, c \in \left(0, \frac{5 \pi}{6}\right)$, we consider two cases:
(i) $a+b+c \geqslant 2 \pi$. In this case,
$a \geqslant 2 \pi - b - c > 2 \pi - \frac{5 \pi}{6} - \frac{5 \pi}{6} = \frac{\pi}{3}$.
Similarly, $b > \frac{\pi}{3}$, $c > \frac{\pi}{3}$.
Thus, $a, b, c \in \left(\frac{\pi}{3}, \frac{5 \pi}{6}\right)$.
Hence, $\sin a, \sin b, \sin c \in \left(\frac{1}{2}, 1\right]$.
Therefore, $\sin a, \sin b, \sin c$ can be the three sides of a triangle.
(ii) $a+b+c < 2 \pi$. In this case, we have $a+b > c$, $b+c > a$, and $c+a > b$.
If $\frac{a+b}{2} \leqslant \frac{\pi}{2}$, since $a+b > c$, then
$0 < \frac{a+b}{2} \leqslant \frac{\pi}{2}$ and $0 < \frac{a-b}{2} < \frac{c}{2} < \frac{\pi}{2}$.
Since $\cos \frac{c}{2} > \cos \frac{5 \pi}{12} > 0$,
$\sin a + \sin b = 2 \sin \frac{a+b}{2} \cdot \cos \frac{a-b}{2} > 2 \sin \frac{c}{2} \cdot \cos \frac{c}{2} = \sin c$.
Similarly, $\sin b + \sin c > \sin a$,
$\sin c + \sin a > \sin b$.
Therefore, $\sin a, \sin b, \sin c$ are also the three sides of a triangle.

In conclusion, when $M=\frac{5 \pi}{6}$, $h(x)=\sin x (x \in (0, M))$ is a triangle-preserving function.
Hence, the maximum value of $M$ is $\frac{5 \pi}{6}$.