Three. (This question is worth 20 points)
Let $x \geqslant y \geqslant z \geqslant \frac{\pi}{12}$, and $x+y+z=\frac{\pi}{2}$, find the maximum and minimum values of the product $\cos x \sin y \cos z$.


$$
\frac{2+\sqrt{3}}{8}
$$

Three, 【Analysis and Solution】From the given conditions, we have
$$
\begin{array}{l}
x=\frac{\pi}{2}-(y+z) \leqslant \frac{\pi}{2}-\left(\frac{\pi}{12}+\frac{\pi}{2}\right)=\frac{\pi}{3}, \\
\sin (x-y) \geqslant 0, \sin (y-z) \geqslant 0 .
\end{array}
$$

Thus, $\cos x \sin y \cos z=\frac{1}{2} \cos x[\sin (y+z)+\sin (y-z)]$
$$
\begin{array}{l}
\geqslant \frac{1}{2} \cos x \sin (y+z) \\
=\frac{1}{2} \cos ^{2} x \geqslant \frac{1}{2} \cos ^{2} \frac{\pi}{3}=\frac{1}{8} .
\end{array}
$$

And when $x=\frac{x}{3}, y=z=\frac{\pi}{2}$, the equality holds.
$\therefore$ The minimum value of $\cos x \sin y \cos z$ is $\frac{1}{8}$.
$$
\begin{array}{l}
\text { Also, } \cos x \sin y \cos z \\
=\frac{1}{2} \cos z[\sin (x+y)-\sin (x-y)] \\
\leqslant \frac{1}{2} \cos z \sin (x+y)=\frac{1}{2} \cos ^{2} z \\
\leqslant \frac{1}{2} \cos ^{2} \frac{\pi}{12}=\frac{1}{4}\left(1+\cos \frac{\pi}{6}\right)=\frac{2+\sqrt{3}}{8} .
\end{array}
$$

And when $x=y=\frac{5 \pi}{24}, z=\frac{\pi}{12}$, the equality holds.
$\therefore$ The maximum value of $\cos x \sin y \cos z$ is $\frac{2+\sqrt{3}}{8}$.