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

$$
\begin{aligned}
& \frac{\tan\left(\frac{5}{4} \pi-4 \alpha\right) \sin ^{2}\left(\frac{5}{4} \pi+4 \alpha\right)}{1-2 \cos ^{2} 4 \alpha}=\frac{\tan\left(\frac{5}{4} \pi-4 \alpha\right) \sin ^{2}\left(\frac{5}{4} \pi+4 \alpha\right)}{-\left(2 \cos ^{2} 4 \alpha-1\right)}= \\
& =\left[\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}, x \neq \pi+2 \pi n, n \in Z\right.
\end{aligned}
$$

$$
\begin{aligned}
& \left.\sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2} ; 2 \cos ^{2} x-1=\cos 2 x\right]= \\
& =\frac{\frac{\sin \left(\frac{5 \pi}{2}-8 \alpha\right)}{1+\cos \left(\frac{5 \pi}{2}-8 \alpha\right)} \cdot \frac{1-\cos \left(\frac{5 \pi}{2}+8 \alpha\right)}{2}}{-\cos 8 \alpha}=\frac{\frac{\cos 8 \alpha}{1+\sin 8 \alpha} \cdot \frac{1+\sin 8 \alpha}{2}}{-\cos 8 \alpha}=-\frac{1}{2} .
\end{aligned}
$$

Answer: $-\frac{1}{2}$.