1. Solve the inequality $2^{\log _{2}^{2} x}-12 \cdot x^{\log _{0.5} x}<3-\log _{3-x}\left(x^{2}-6 x+9\right)$.
Solution.

$$
\begin{gathered}
2^{\log _{2}^{2} x}-12 \cdot x^{\log _{0.5} x} < 0. \text{ Let } y = x^{\log _{2} x}. \text{ Then (1) } \Leftrightarrow y-\frac{12}{y} < 0, \text{ so } y \in (0,4).
\end{gathered}
$$

From here, $x^{\log _{2} x} < 4 \Leftrightarrow \log _{2}^{2} x < 2 \Leftrightarrow -\sqrt{2} < \log _{2} x < \sqrt{2} \Leftrightarrow 2^{-\sqrt{2}} < x < 2^{\sqrt{2}}$ - this is the answer to inequality (1). Next, consider the restrictions (2). For this, compare the numbers $2^{\sqrt{2}}$ and 3.

Note that $2^{\sqrt{2}} < 2^{1.5}$ and $2^{1.5} < 3$, since $8 = (2^{1.5})^2 < 3^2 = 9$. Therefore, $2^{\sqrt{2}} < 3$.

Write the answer taking (2) into account.

Answer: $x \in \left(2^{-\sqrt{2}}, 2\right) \cup \left(2, 2^{\sqrt{2}}\right)$.