3.312. $2 \cos ^{2} 2 \alpha+\sqrt{3} \sin 4 \alpha-1$.

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3.312. $2 \cos ^{2} 2 \alpha+\sqrt{3} \sin 4 \alpha-1$.
Solution.

$$
\begin{aligned}
& 2 \cos ^{2} 2 \alpha+\sqrt{3} \sin 4 \alpha-1=\left[\cos ^{2} \frac{x}{2}=\frac{1+\cos x}{2}\right]= \\
& =1+\cos 4 \alpha+\sqrt{3} \sin 4 \alpha-1= \\
& =\cos 4 \alpha+\sqrt{3} \sin 4 \alpha=2\left(\frac{1}{2} \cos 4 \alpha+\frac{\sqrt{3}}{2} \sin 4 \alpha\right)= \\
& =2\left(\cos \frac{\pi}{3}+\sin \frac{\pi}{3} \sin 4 \alpha\right)=[\cos x \cos y+\sin x \sin y=\cos (x-y)]= \\
& =2 \cos \left(\frac{\pi}{3}-4 \alpha\right) \\
& \text { Answer: } 2 \cos \left(\frac{\pi}{3}-4 \alpha\right)
\end{aligned}
$$

$$
\text { 3.313. } \frac{\sin \left(\frac{9}{2} \pi-2 \alpha\right)+2 \sin ^{2}\left(2 \alpha-\frac{5}{2} \pi\right)-1}{1+\sin \left(2 \alpha+\frac{\pi}{2}\right)-\sin \left(4 \alpha-\frac{\pi}{2}\right)+\sin \left(6 \alpha-\frac{3}{2} \pi\right)}
$$

Solution.

$$
\begin{aligned}
& \frac{\sin \left(\frac{9}{2} \pi-2 \alpha\right)+2 \sin ^{2}\left(2 \alpha-\frac{5}{2} \pi\right)-1}{1+\sin \left(2 \alpha+\frac{\pi}{2}\right)-\sin \left(4 \alpha-\frac{\pi}{2}\right)+\sin \left(6 \alpha-\frac{3}{2} \pi\right)}= \\
& =\frac{\sin \left(4 \pi+\left(\frac{\pi}{2}-2 \alpha\right)\right)+2 \sin ^{2}\left(\frac{5}{2} \pi-2 \alpha\right)-1}{1+\sin \left(\frac{\pi}{2}+2 \alpha\right)+\sin \left(\frac{\pi}{2}-4 \alpha\right)-\sin \left(\frac{3}{2} \pi-6 \alpha\right)}= \\
& =\frac{\sin \left(4 \pi+\left(\frac{\pi}{2}-2 \alpha\right)\right)-\left(1-2 \sin ^{2}\left(\frac{5}{2} \pi-2 \alpha\right)\right)}{1+\sin \left(\frac{\pi}{2}+2 \alpha\right)+\sin \left(\frac{\pi}{2}-4 \alpha\right)-\sin \left(\frac{3}{2} \pi-6 \alpha\right)}= \\
& =\frac{\cos 2 \alpha-\cos (5 \pi-4 \alpha)}{1+\cos 2 \alpha+\cos 4 \alpha+\cos 6 \alpha}=\frac{\cos 2 \alpha+\cos 4 \alpha}{1+\cos 2 \alpha+\cos 4 \alpha+\cos 6 \alpha}= \\
& =\frac{\cos 2 \alpha+\cos 4 \alpha}{(1+\cos 4 \alpha)+(\cos 6 \alpha+\cos 2 \alpha)}=[\cos 2 x=2 \cos x-1
\end{aligned}
$$

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

Answer: $\frac{1}{2 \cos 2 \alpha}$.