$\sin ^{4} \frac{\pi}{8}+\sin ^{4} \frac{3 \pi}{8}+\sin ^{4} \frac{5 \pi}{8}+\sin ^{4} \frac{7 \pi}{8}$

$=\frac{1}{4}\left[\left(2 \sin ^{2} \frac{\pi}{8}\right)^{2}+\left(2 \sin ^{2} \frac{3 \pi}{8}\right)^{2}\right]$

$\quad+\frac{1}{4}\left[\left(2 \sin ^{2} \frac{\pi}{8}\right)^{2}+\left(2 \sin ^{2} \frac{3 \pi}{8}\right)^{2}\right]$

$=\frac{1}{4}\left[\left(1-\cos \frac{\pi}{4}\right)^{2}+\left(1-\cos \frac{3 \pi}{4}\right)^{2}\right]$

$\quad+\frac{1}{4}\left[\left(1-\cos \frac{\pi}{4}\right)^{2}+\left(1-\cos \frac{3 \pi}{4}\right)^{2}\right]$

$=\frac{1}{4}\left[\left(1-\frac{1}{\sqrt{2}}\right)^{2}+\left(1+\frac{1}{\sqrt{2}}\right)^{2}\right]$

$+\frac{1}{4}\left[\left(1-\frac{1}{\sqrt{2}}\right)^{2}+\left(1+\frac{1}{\sqrt{2}}\right)^{2}\right]$

$=\frac{1}{4}(3)+\frac{1}{4}(3)=\frac{3}{2}$