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

Let $S = \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \cos^4 \frac{7\pi}{8}$.

Observe that $\cos \frac{5\pi}{8} = -\cos \frac{3\pi}{8}$ and $\cos \frac{7\pi}{8} = -\cos \frac{\pi}{8}$, but the fourth powers eliminate the negative signs: $\cos^4 \frac{5\pi}{8} = \cos^4 \frac{3\pi}{8}$ and $\cos^4 \frac{7\pi}{8} = \cos^4 \frac{\pi}{8}$.

Thus, $S = 2\left(\cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8}\right)$.

Since $\cos \frac{3\pi}{8} = \sin \frac{\pi}{8}$, we have $S = 2\left(\cos^4 \frac{\pi}{8} + \sin^4 \frac{\pi}{8}\right)$.

Using the identity $a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2$ with $a = \cos \frac{\pi}{8}$, $b = \sin \frac{\pi}{8}$, and noting that $\sin^2 x + \cos^2 x = 1$:

$S = 2\left(1 - 2\cos^2 \frac{\pi}{8} \sin^2 \frac{\pi}{8}\right)$.

From the double-angle identity, $2\sin^2 x \cos^2 x = \frac{1}{2}\sin^2 2x$, so:

$S = 2\left(1 - \frac{1}{2}\sin^2 \frac{\pi}{4}\right)$.

Since $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, we have $\sin^2 \frac{\pi}{4} = \frac{1}{2}$, and thus:

$S = 2\left(1 - \frac{1}{4}\right) = 2 \cdot \frac{3}{4} = \frac{3}{2}$.

Final answer: $\boxed{\frac{3}{2}}$