3.308. $\cos ^{2}(\alpha-2 \beta)-\cos ^{2}\left(\alpha-\frac{\pi}{2}\right)-\cos ^{2}(2 \beta-\pi)$.
Solution.

$$
\begin{aligned}
& \cos ^{2}(\alpha-2 \beta)-\cos ^{2}\left(\alpha-\frac{\pi}{2}\right)-\cos ^{2}(2 \beta-\pi)= \\
& =\cos ^{2}(\alpha-2 \beta)-\cos ^{2}\left(\frac{\pi}{2}-\alpha\right)-\cos ^{2}(\pi-2 \beta)=\left[\cos ^{2} \frac{x}{2}=\frac{1+\cos x}{2}\right]= \\
& =\frac{1+\cos (2 \alpha-4 \beta)}{2}-\frac{1+\cos (\pi-2 \alpha)}{2}-\frac{1+\cos (2 \pi-4 \beta)}{2}= \\
& =\frac{1}{2}(\cos (2 \alpha-4 \beta)-\cos (\pi-2 \alpha)-\cos (2 \pi-4 \beta)-1)= \\
& =\frac{1}{2}(\cos (2 \alpha-4 \beta)+\cos 2 \alpha-\cos 4 \beta-1)= \\
& =\left[\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} ; \cos 2 x=2 \cos ^{2} x-1\right]= \\
& =\frac{1}{2}(2 \cos (2 \alpha-2 \beta) \cos 2 \beta-2 \cos 2 \beta)=\cos 2 \beta(\cos (2 \alpha-2 \beta)-\cos 2 \beta)= \\
& =\left[\cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right]=\cos 2 \beta(-2 \sin \alpha \sin (\alpha-2 \beta))= \\
& =2 \sin \alpha \sin (2 \beta-\alpha) \cos 2 \beta
\end{aligned}
$$

Answer: $2 \sin \alpha \sin (2 \beta-\alpha) \cos 2 \beta$.