3.277. $\sin ^{2}\left(135^{\circ}-2 \alpha\right)-\sin ^{2}\left(210^{\circ}-2 \alpha\right)-\sin 195^{\circ} \cos \left(165^{\circ}-4 \alpha\right)$.
Solution.

$$
\begin{aligned}
& \sin ^{2}\left(135^{\circ}-2 \alpha\right)-\sin ^{2}\left(210^{\circ}-2 \alpha\right)-\sin 195^{\circ} \cos \left(165^{\circ}-4 \alpha\right)= \\
& =\left(\sin \left(180^{\circ}-\left(45^{\circ}+2 \alpha\right)\right)\right)^{2}-\left(\sin \left(180^{\circ}+\left(30^{\circ}-2 \alpha\right)\right)\right)^{2}-\sin \left(180^{\circ}+15^{\circ}\right) \times \\
& \times \cos \left(180^{\circ}-\left(15^{\circ}+4 \alpha\right)\right)= \\
& =\sin ^{2}\left(45^{\circ}+2 \alpha\right)-\sin ^{2}\left(30^{\circ}-2 \alpha\right)-\left(-\sin 15^{\circ}\right)\left(-\cos \left(15^{\circ}+4 \alpha\right)\right)= \\
& =\sin ^{2}\left(45^{\circ}+2 \alpha\right)-\sin ^{2}\left(30^{\circ}-2 \alpha\right)-\sin 15^{\circ} \cos \left(15^{\circ}+4 \alpha\right)= \\
& =\left[\sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2} ; \sin x \cos y=\frac{1}{2}(\sin (x-y)+\sin (x+y))\right]= \\
& =\frac{1-\cos \left(90^{\circ}+4 \alpha\right)}{2}-\frac{1-\cos \left(60^{\circ}-4 \alpha\right)}{2}-\frac{1}{2}\left(\sin (-4 \alpha)+\sin \left(30^{\circ}+4 \alpha\right)\right)= \\
& =\frac{1}{2}\left(1-\cos \left(90^{\circ}+4 \alpha\right)-1+\cos \left(60^{\circ}-4 \alpha\right)+\sin 4 \alpha-\sin \left(30^{\circ}+4 \alpha\right)\right)= \\
& =\frac{1}{2}\left(\sin 4 \alpha+\cos \left(60^{\circ}-4 \alpha\right)+\sin 4 \alpha-\sin \left(90^{\circ}-\left(60^{\circ}-4 \alpha\right)\right)\right)= \\
& =\frac{1}{2}\left(2 \sin 4 \alpha+\cos \left(60^{\circ}-4 \alpha\right)-\cos \left(60^{\circ}-4 \alpha\right)\right)=\sin 4 \alpha .
\end{aligned}
$$

Answer: $\sin 4 \alpha$.