3.111.

$$
\frac{\cos ^{2}\left(\frac{5}{2} \pi-2 \alpha\right)+4 \cos ^{2}\left(\frac{7}{2} \pi-\alpha\right)-4}{1+\cos (4 \alpha-\pi)-8 \sin ^{2}(5 \pi-\alpha)}
$$
Solution.

$$
\begin{aligned}
& \frac{\cos ^{2}\left(\frac{5}{2} \pi-2 \alpha\right)+4 \cos ^{2}\left(\frac{7}{2} \pi-\alpha\right)-4}{1+\cos (4 \alpha-\pi)-8 \sin ^{2}(5 \pi-\alpha)}= \\
& =\frac{\left(\cos \left(\frac{4 \pi+\pi}{2}-2 \alpha\right)\right)^{2}+4\left(\cos \left(\frac{6 \pi+\pi}{2}-\alpha\right)\right)^{2}-4}{1+\cos (\pi-4 \alpha)-8(\sin (5 \pi-\alpha))^{2}}= \\
& =\frac{\left(\cos \left(2 \pi+\left(\frac{\pi}{2}-2 \alpha\right)\right)\right)^{2}+4\left(\cos \left(3 \pi+\left(\frac{\pi}{2}-\alpha\right)\right)\right)^{2}-4}{1+\cos (\pi-4 \alpha)-8(\sin (5 \pi-\alpha))^{2}}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{\left(\cos \left(\frac{\pi}{2}-2 \alpha\right)\right)^{2}+4\left(-\cos \left(\frac{\pi}{2}-\alpha\right)\right)^{2}-4}{1-\cos 4 \alpha-8 \sin ^{2} \alpha}=\frac{\sin ^{2} 2 \alpha+4 \sin ^{2} \alpha-4}{1-\cos 4 \alpha-8 \sin ^{2} \alpha}= \\
& =\frac{4 \sin ^{2} \alpha \cos ^{2} \alpha-4\left(1-\sin ^{2} \alpha\right)}{1-\left(1-2 \sin ^{2} 2 \alpha\right)-8 \sin ^{2} \alpha}=\frac{4 \sin ^{2} \alpha \cos ^{2} \alpha-4 \cos ^{2} \alpha}{2 \sin ^{2} 2 \alpha-8 \sin ^{2} \alpha}= \\
& =\frac{4 \cos ^{2} \alpha\left(\sin ^{2} \alpha-1\right)}{8 \sin ^{2} \alpha \cos ^{2} \alpha-8 \sin ^{2} \alpha}=\frac{-4 \cos ^{2} \alpha \cos ^{2} \alpha}{8 \sin ^{2} \alpha\left(\cos ^{2} \alpha-1\right)}=\frac{-4 \cos ^{4} \alpha}{-8 \sin ^{2} \alpha \sin ^{2} \alpha}= \\
& =\frac{\cos ^{4} \alpha}{2 \sin ^{4} \alpha}=\frac{1}{2} \operatorname{ctg}^{4} \alpha
\end{aligned}
$$

Answer: $\frac{1}{2} \operatorname{ctg}^{4} \alpha$.