3.321. $\cos ^{2}\left(\frac{5}{8} \pi+\alpha\right)-\sin ^{2}\left(\frac{15}{8} \pi+\alpha\right)$.
Solution.

$$
\begin{aligned}
& \cos ^{2}\left(\frac{5}{8} \pi+\alpha\right)-\sin ^{2}\left(\frac{15}{8} \pi+\alpha\right)= \\
& =\left[\cos ^{2} \frac{x}{2}=\frac{1+\cos x}{2}, \sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2}\right]= \\
& =\frac{1+\cos \left(\frac{5 \pi}{4}+2 \alpha\right)}{2}-\frac{1-\cos \left(\frac{15 \pi}{4}+2 \alpha\right)}{2}= \\
& =\frac{1}{2}\left(\cos \left(\frac{5 \pi}{4}+2 \alpha\right)+\cos \left(\frac{15 \pi}{4}+2 \alpha\right)\right)= \\
& =\frac{1}{2}\left(\cos \left(\pi+\left(\frac{\pi}{4}+2 \alpha\right)\right)+\cos \left(4 \pi-\left(\frac{\pi}{4}-2 \alpha\right)\right)\right)= \\
& =\frac{1}{2}\left(-\cos \left(\frac{\pi}{4}+2 \alpha\right)+\cos \left(\frac{\pi}{4}-2 \alpha\right)\right)= \\
& =\frac{1}{2}\left(\cos \left(\frac{\pi}{4}-2 \alpha\right)-\cos \left(\frac{\pi}{4}+2 \alpha\right)\right)= \\
& =\left[\cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right]=\frac{1}{2}\left(-2 \sin \frac{\pi}{4} \sin (-2 \alpha)\right)= \\
& =\sin \frac{\pi}{4} \sin 2 \alpha=\frac{\sqrt{2}}{2} \sin 2 \alpha .
\end{aligned}
$$

Answer: $\frac{\sqrt{2}}{2} \sin 2 \alpha$.