3.299. $2-\frac{\sin 8 \alpha}{\sin ^{4} 2 \alpha-\cos ^{4} 2 \alpha}$.
## Solution.

$$
\begin{aligned}
& 2-\frac{\sin 8 \alpha}{\sin ^{4} 2 \alpha-\cos ^{4} 2 \alpha}=2+\frac{\sin 2(4 \alpha)}{\cos ^{4} 2 \alpha-\sin ^{4} 2 \alpha}= \\
& =2+\frac{2 \sin 4 \alpha \cos 4 \alpha}{\left(\cos ^{2} 2 \alpha-\sin ^{2} 2 \alpha\right)\left(\cos ^{2} 2 \alpha+\sin ^{2} 2 \alpha\right)}=2+\frac{2 \sin 4 \alpha \cos 4 \alpha}{\cos ^{2} 2 \alpha-\sin ^{2} 2 \alpha}= \\
& =2+\frac{2 \sin 4 \alpha \cos 4 \alpha}{\cos 4 \alpha}=2+2 \sin 4 \alpha=2(1+\sin 4 \alpha)= \\
& =2\left(\cos ^{2} 2 \alpha+2 \cos 2 \alpha \sin 2 \alpha+\sin ^{2} 2 \alpha\right)=2(\cos 2 \alpha+\sin 2 \alpha)^{2}= \\
& =2\left(\frac{2}{\sqrt{2}}\right)^{2}\left(\frac{\sqrt{2}}{2} \cos 2 \alpha+\frac{\sqrt{2}}{2} \sin 2 \alpha\right)^{2}=4\left(\cos \frac{\pi}{4} \cos 2 \alpha+\sin \frac{\pi}{4} \sin 2 \alpha\right)^{2}= \\
& =[\cos x \cos y+\sin x \sin y=\cos (x-y)]= \\
& =4\left(\cos \left(\frac{\pi}{4}-2 \alpha\right)\right)^{2}=4 \cos ^{2}\left(\frac{\pi}{4}-2 \alpha\right)
\end{aligned}
$$

Answer: \(4 \cos ^{2}\left(\frac{\pi}{4}-2 \alpha\right)\).