3.258. $\cos ^{-2} 4 \alpha-\operatorname{tg}^{2}(3 \pi+4 \alpha)-2 \cos ^{2} \alpha-\sqrt{3} \cos \left(\frac{3}{2} \pi-2 \alpha\right)$.
Solution.

$$
\cos ^{-2} 4 \alpha-\operatorname{tg}^{2}(3 \pi+4 \alpha)-2 \cos ^{2} \alpha-\sqrt{3} \cos \left(\frac{3}{2} \pi-2 \alpha\right)=
$$

$$
\begin{aligned}
& =\frac{1}{\cos ^{2} 4 \alpha}-(\operatorname{tg}(3 \pi+4 \alpha))^{2}-2 \cos ^{2} \alpha-\sqrt{3} \cos \left(\frac{3}{2} \pi-2 \alpha\right)= \\
& =\frac{1}{\cos ^{2} 4 \alpha}-\operatorname{tg}^{2} 4 \alpha-2 \cos ^{2} 2 \alpha+\sqrt{3} \sin 2 \alpha= \\
& =\frac{1}{\cos ^{2} 4 \alpha}-\frac{\sin ^{2} 4 \alpha}{\cos ^{2} 4 \alpha}-2 \cos ^{2} \alpha+\sqrt{3} \sin 2 \alpha= \\
& =\frac{1-\sin ^{2} 4 \alpha}{\cos ^{2} 4 \alpha}-2 \cos ^{2} \alpha+\sqrt{3} \sin 2 \alpha=\frac{\cos ^{2} 4 \alpha}{\cos ^{2} 4 \alpha}-2 \cos ^{2} \alpha+\sqrt{3} \sin 2 \alpha= \\
& =1-2 \cos ^{2} \alpha+\sqrt{3} \sin 2 \alpha=-\left(2 \cos ^{2} \alpha-1\right)+\sqrt{3} \sin 2 \alpha= \\
& =\sqrt{3} \sin 2 \alpha-\cos 2 \alpha=2\left(\frac{\sqrt{3}}{2} \sin 2 \alpha-\frac{1}{2} \cos 2 \alpha\right)= \\
& =2\left(\cos \frac{\pi}{6} \sin 2 \alpha-\sin \frac{\pi}{6} \cos 2 \alpha\right)= \\
& =2\left(\sin 2 \alpha \cos \frac{\pi}{6}-\cos 2 \alpha \sin \frac{\pi}{6}\right)=2 \sin \left(2 \alpha-\frac{\pi}{6}\right)
\end{aligned}
$$

Answer: $2 \sin \left(2 \alpha-\frac{\pi}{6}\right)$.