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

$$
\begin{aligned}
& \frac{\cos ^{2}(4 \alpha-3 \pi)-4 \cos ^{2}(2 \alpha-\pi)+3}{\cos ^{2}(4 \alpha+3 \pi)+4 \cos ^{2}(2 \alpha+\pi)-1}= \\
& =\frac{(\cos (3 \pi-4 \alpha))^{2}-4(\cos (\pi-2 \alpha))^{2}+3}{(\cos (3 \pi+4 \alpha))^{2}+4(\cos (\pi+2 \alpha))^{2}-1}=\frac{\cos ^{2} 4 \alpha-4 \cos ^{2} 2 \alpha+3}{\cos ^{2} 4 \alpha+4 \cos ^{2} 2 \alpha-1}=
\end{aligned}
$$

$=\frac{(\cos 2(2 \alpha))^{2}-4\left(1-\sin ^{2} 2 \alpha\right)+3}{(\cos 2(2 \alpha))^{2}+4 \cos ^{2} 2 \alpha-1}=\frac{\left(1-2 \sin ^{2} 2 \alpha\right)^{2}-4+4 \sin ^{2} 2 \alpha+3}{\left(2 \cos ^{2} 2 \alpha-1\right)^{2}+4 \cos ^{2} 2 \alpha-1}=$

$=\frac{1-4 \sin ^{2} 2 \alpha+4 \sin ^{4} 2 \alpha-4+4 \sin ^{2} 2 \alpha+3}{4 \cos ^{4} 2 \alpha-4 \cos ^{2} 2 \alpha+1+4 \cos ^{2} 2 \alpha-1}=\frac{4 \sin ^{4} 2 \alpha}{4 \cos ^{4} 2 \alpha}=\frac{\sin ^{4} 2 \alpha}{\cos ^{4} 2 \alpha}=$

$=\operatorname{tg}^{4} 2 \alpha$.

Answer: $\operatorname{tg}^{4} 2 \alpha$.