13. From the conditions, we know that $x=\frac{\pi}{2}-(y+z) \leqslant \frac{\pi}{2}-\left(\frac{\pi}{12}+\frac{\pi}{12}\right)=\frac{\pi}{3}, \sin (x-y) \geqslant 0, \sin (y-$ $z) \geqslant 0$, thus $\cos x \sin y \cos z=\frac{1}{2} \cos x[\sin (y+z)+\sin (y-z)] \geqslant \frac{1}{2} \cos x \sin (y+z)=$ $\frac{1}{2} \cos ^{2} x \geqslant \frac{1}{8}$. When $x=\frac{11}{3}, y=z=\frac{11}{12}$, the equality holds, and the minimum value of $\cos x \sin y \cos z$ is $\frac{1}{8}$. Also, $\cos x \sin y \cos z=\frac{1}{2} \cos z[\sin (x+y)-\sin (x-y)] \leqslant \frac{1}{2} \cos z \sin (x+y) \leqslant \frac{1}{2} \cos ^{2} z \leq$ $\frac{1}{2} \cos ^{2} \frac{\pi}{12}=\frac{1}{4}+\frac{\sqrt{3}}{8}$, when $x=y=\frac{5}{24} \pi, z=\frac{\pi}{12}$, the equality holds, and the maximum value of $\cos x \sin y \cos z$ is $\frac{1}{4}+$ $\frac{\sqrt{3}}{8}$.