9.212. $\left\{\begin{array}{l}\left(\frac{2}{3}\right)^{x} \cdot\left(\frac{8}{9}\right)^{-x}>\frac{27}{64} \\ 2^{x^{2}-6 x-3.5}<8 \sqrt{2}\end{array}\right.$
## Solution.

Let's rewrite the given system of inequalities as

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ \frac { 2 ^ { x - 3 x } } { 3 ^ { x - 2 x } } > \frac { 27 } { 64 } , } \\
{ 2 ^ { x ^ { 2 } - 6 x - 3.5 } \left(\frac{3}{4}\right)^{3}, \\
x^{2}-6 x-3.5<3.5
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x < 3 , } \\
{ x ^ { 2 } - 6 x - 7 < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x<3, \\
-1<x<7 .
\end{array}\right.\right.
\end{aligned}
$$

Answer: $x \in(-1 ; 3)$.