1. Let $A=\sin \left(\sin \frac{3}{8} \pi\right), B=\sin \left(\cos \frac{3}{8} \pi\right), C=\cos \left(\sin \frac{3}{8} \pi\right), D=$ $\cos \left(\cos \frac{3}{8} \pi\right)$, compare the sizes of $A$, $B$, $C$, and $D$.
1. From $\sin \frac{\pi}{4}<\sin \left(\sin \frac{3}{8} \pi\right)<\sin \frac{\pi}{3}, \cos \frac{\pi}{3}<\cos \left(\sin \frac{3}{8} \pi\right)<\cos \frac{\pi}{4}$, we get $\frac{1}{2}<C<\frac{\sqrt{2}}{2}<A<\frac{\sqrt{3}}{2}$.
From $\sin 0<\sin \left(\cos \frac{3}{8} \pi\right)<\sin \frac{\pi}{6}, \cos \frac{\pi}{6}<\cos \left(\cos \frac{3}{8} \pi\right)<\cos 0$, we can deduce $0<B<\frac{1}{2}, \quad \frac{\sqrt{3}}{2}<D<1$. Therefore, $B<C<A<D$.