$8.57 \log _{3} \log _{4} \frac{4 x-1}{x+1}-\log _{\frac{1}{3}} \log _{\frac{1}{4}} \frac{x+1}{4 x-1}<0$.
8.57 Noticing that

$$
\begin{aligned}
& \log _{\frac{1}{4}} \frac{x+1}{4 x-1}=\log _{4} \frac{4 x-1}{x+1} \\
& \log _{\frac{1}{3}} \log _{4} \frac{4 x-1}{x+1}=-\log _{3} \log _{4} \frac{4 x-1}{x+1}
\end{aligned}
$$

we transform the given inequality:

$$
\begin{aligned}
& \log _{3} \log _{4} \frac{4 x-1}{x+1}+\log _{3} \log _{4} \frac{4 x-1}{x+1}>\frac{2}{3}.
\end{aligned}
$$

Answer: $\left(\frac{2}{3}, \infty\right)$.