8.275. $\frac{\cot 4 t}{\sin ^{2} t}+\frac{\cot t}{\sin ^{2} 4 t}=0$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\sin t \neq 0, \\ \sin 4 t \neq 0 .\end{array}\right.$

Rewrite the equation as

$$
\frac{\frac{\cos 4 t}{\sin 4 t}}{\sin ^{2} t}+\frac{\frac{\cos t}{\sin t}}{\sin ^{2} 4 t}=0, \frac{\cos 4 t}{\sin 4 t \sin ^{2} t}+\frac{\cos t}{\sin t \sin ^{2} 4 t}=0 \Leftrightarrow
$$

$\Leftrightarrow \sin 4 t \cos 4 t+\sin t \cos t=0 \Leftrightarrow \frac{\sin 8 t}{2}+\frac{\sin 2 t}{2}=0 \Leftrightarrow$ $\Leftrightarrow \sin 8 t+\sin 2 t=0 \Leftrightarrow 2 \sin 5 t \cos 3 t=0$.

From this,

1) $\sin 5 t=0,5 t=\pi k, t_{1}=\frac{\pi k}{5}, k \in Z$;
2) $\cos 3 t=0,3 t=\frac{\pi}{2}+\pi n, t_{2}=\frac{\pi}{6}+\frac{\pi n}{3}=\frac{\pi}{6}(2 n+1), n \in Z$.

Considering the domain of definition, we get

$$
t_{1}=\frac{\pi k}{5}, k \neq 5 l ; t_{2}=\frac{\pi}{6}+\frac{\pi n}{3}=\frac{\pi}{6}(2 n+1), n \neq 3 l+1, l \in Z
$$

Answer: $t_{1}=\frac{\pi k}{5}, k \neq 5 l ; t_{2}=\frac{\pi}{6}(2 n+1), n \neq 3 l+1$, where $k, n$ and $l \in Z$.