28. Given $x \geqslant y \geqslant z \geqslant \frac{\pi}{12}$, and $x+y+z=\frac{\pi}{2}$, find the maximum and minimum values of $\cos x \sin y \cos z$.
28. $\cos x \sin y \cos z=\cos x \cdot \frac{\sin (y+z)+\sin (y-z)}{2} \geqslant \frac{1}{2} \cos x \sin (y-z)=\frac{1}{2} \cos ^{2} x$.
Since $x=\frac{\pi}{2}-(y+z) \leqslant \frac{\pi}{2}-2 \cdot \frac{\pi}{12}=\frac{\pi}{3}$.
Therefore, $\cos x \geqslant \cos \frac{\pi}{3}=\frac{1}{2}$.
Thus, $\cos x \sin y \cos z \geqslant \frac{1}{2} \cos ^{2} x \geqslant \frac{1}{2} \cos ^{2} \frac{\pi}{3}=\frac{1}{8}$.
Equality holds if and only if $x=\frac{\pi}{3}, y=z=\frac{\pi}{12}$.
$$
\begin{aligned}
\cos x \sin y \cos z & =\cos z \cdot \frac{\sin (y+x)+\sin (y-x)}{2}=\cos z \cdot \frac{\sin (x+y)-\sin (x-y)}{2} \\
& \leqslant \frac{1}{2} \cos z \sin (x+y) \\
& =\frac{1}{2} \cos z \cdot \sin \left(\frac{\pi}{2}-z\right)=\frac{1}{2} \cos ^{2} z=\frac{1+\cos 2 z}{4} \\
& \leqslant \frac{1+\cos 2 \cdot \frac{\pi}{12}}{4}=\frac{2+\sqrt{3}}{8} .
\end{aligned}
$$
Equality holds if and only if $x=y=\frac{5 \pi}{24}, z=\frac{\pi}{12}$.
In summary, the minimum value of $\cos x \sin y \cos z$ is $\frac{1}{8}$, and the maximum value is $\frac{2+\sqrt{3}}{8}$.