1. **Problem statement:** We want to prove that $$\cos^4\left(\frac{\pi}{8}\right) + \cos^4\left(\frac{3\pi}{8}\right) + \cos^4\left(\frac{5\pi}{8}\right) + \cos^4\left(\frac{7\pi}{8}\right) = \frac{3}{2}.$$\n\n2. **Recall the power reduction formula:** \nWe can use the identity $$\cos^4 \theta = \left(\cos^2 \theta\right)^2 = \left(\frac{1 + \cos 2\theta}{2}\right)^2 = \frac{1}{4}\left(1 + 2\cos 2\theta + \cos^2 2\theta\right).$$\n\n3. **Express \(\cos^2 2\theta\) using the power reduction formula again:**\n$$\cos^2 2\theta = \frac{1 + \cos 4\theta}{2}.$$\nSubstitute back to get\n$$\cos^4 \theta = \frac{1}{4}\left(1 + 2\cos 2\theta + \frac{1 + \cos 4\theta}{2}\right) = \frac{1}{4}\left(1 + 2\cos 2\theta + \frac{1}{2} + \frac{\cos 4\theta}{2}\right) = \frac{3}{8} + \frac{1}{2}\cos 2\theta + \frac{1}{8}\cos 4\theta.$$\n\n4. **Calculate the sum:**\nLet $$S = \sum_{k=1}^4 \cos^4 \theta_k$$ where $$\theta_1 = \frac{\pi}{8}, \theta_2 = \frac{3\pi}{8}, \theta_3 = \frac{5\pi}{8}, \theta_4 = \frac{7\pi}{8}.$$\nUsing the expression above for each term,\n$$S = \sum_{k=1}^4 \left(\frac{3}{8} + \frac{1}{2} \cos 2\theta_k + \frac{1}{8} \cos 4\theta_k\right) = 4 \cdot \frac{3}{8} + \frac{1}{2} \sum_{k=1}^4 \cos 2\theta_k + \frac{1}{8} \sum_{k=1}^4 \cos 4\theta_k.$$\n\n5. **Evaluate the cosine sums:**\nNote that $$\{2\theta_k\} = \left\{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\right\}$$ and $$\{4\theta_k\} = \left\{\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}\right\}.$$\nSum of cosines at quadruple angles:\n- $$\sum \cos 2\theta_k = \cos \frac{\pi}{4} + \cos \frac{3\pi}{4} + \cos \frac{5\pi}{4} + \cos \frac{7\pi}{4} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0.$$\n- $$\sum \cos 4\theta_k = \cos \frac{\pi}{2} + \cos \frac{3\pi}{2} + \cos \frac{5\pi}{2} + \cos \frac{7\pi}{2} = 0 + 0 + 0 + 0 = 0.$$\n\n6. **Substitute the sums back into S:**\n$$S = 4 \cdot \frac{3}{8} + \frac{1}{2} \cdot 0 + \frac{1}{8} \cdot 0 = \frac{12}{8} = \frac{3}{2}.$$\n\n**Answer:** $$\boxed{\frac{3}{2}}.$$