3.253.

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

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

Answer: $\tan \alpha$.