Five, (16 points) When $x \leqslant y \leqslant z$, find the positive integer solutions to the equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{7}{8}$.
First, given $x>1$, and $\frac{1}{x} \geqslant \frac{1}{y} \geqslant \frac{1}{z}>0$. Therefore,
$$
\begin{array}{l}
\frac{1}{x}<\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{3}{x} \\
\Rightarrow \frac{1}{x}<\frac{7}{8} \leqslant \frac{3}{x} \\
\Rightarrow \frac{8}{7}<x \leqslant \frac{24}{7} .
\end{array}
$$

Thus, $x=2$ or 3.
If $x=2$, then $\frac{1}{y}+\frac{1}{z}=\frac{3}{8}$.
Hence $\frac{1}{y}<\frac{1}{y}+\frac{1}{z} \leqslant \frac{2}{y}$
$\Rightarrow \frac{1}{y}<\frac{3}{8} \leqslant \frac{2}{y}$
$\Rightarrow \frac{8}{3}<y \leqslant \frac{16}{3}$.
Thus, $y=3,4,5$.
From $\frac{1}{y}+\frac{1}{z}=\frac{3}{8}$, we get $z=24,8, \frac{40}{7}$ (discard $\frac{40}{7}$).
If $x=3$, then $\frac{1}{y}+\frac{1}{z}=\frac{13}{24}$.
Hence $\frac{1}{y}<\frac{1}{y}+\frac{1}{z} \leqslant \frac{2}{y}$
$\Rightarrow \frac{1}{y}<\frac{13}{24} \leqslant \frac{2}{y}$
$\Rightarrow \frac{24}{13}<y \leqslant \frac{48}{13}$.
Thus, $y=2,3$ (discard $y=2$).
From $\frac{1}{y}+\frac{1}{z}=\frac{13}{24}$, we get $z=\frac{24}{5}$ (discard $\frac{24}{5}$).
Therefore, the positive integer solutions $(x, y, z)$ of the equation are
$(2,3,24)$ and $(2,4,8)$.