3.106.

$$
\frac{1+\cos (4 \alpha-2 \pi)+\cos \left(4 \alpha-\frac{\pi}{2}\right)}{1+\cos (4 \alpha+\pi)+\cos \left(4 \alpha+\frac{3}{2} \pi\right)}
$$
Solution.

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

Answer: $\operatorname{ctg} 2 \alpha$.