$8.55 \quad \log _{\frac{1}{3}} \log _{\frac{1}{2}} \frac{x+4}{2 x-3}<0$.

$8.55 \quad \log _{\frac{1}{3}} \log _{\frac{1}{2}} \frac{x+4}{2 x-3}<0$.
8.55 Replacing 0 with $\log _{\frac{1}{3}} 1$ and considering that the base of the first logarithm satisfies the inequality $0<\log _{\frac{1}{3}} 1<1$ (in this case, the condition $\log _{\frac{1}{2}} \frac{x+4}{2 x-3}>0$ is automatically satisfied). Further, since $1=\log _{\frac{1}{2}} \frac{1}{2}$ and $0<\log _{\frac{1}{2}} \frac{x+4}{2 x-3}<1$, we have:

\[
\left\{
\begin{array}{l}
\frac{x+4}{2 x-3} > 0, \\
\frac{1}{2} < \frac{x+4}{2 x-3} < 1.
\end{array}
\right.
\]

From the second inequality of the system, it follows that $2 x - 3 < 0$; hence, $x + 4 < 0$ and the problem reduces to solving the equivalent system

\[
\left\{
\begin{array}{l}
2 x - 3 < 0, \\
x + 4 < 0,
\end{array}
\right.
\]
or
\[
\left\{
\begin{array}{l}
x < \frac{3}{2}, \\
x < -4,
\end{array}
\right.
\]

from which $x < -4$.

Answer: $(-\infty, -4)$.