$7.45 \sqrt[4]{|x-3|^{x+1}}=\sqrt[3]{|x-3|^{x-2}}$.

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$7.45 \sqrt[4]{|x-3|^{x+1}}=\sqrt[3]{|x-3|^{x-2}}$.
7.45 We have

$|x-3|^{\frac{x+1}{4}}=|x-3|^{\frac{x-2}{3}}$

According to the hint $2^{0}$, the roots of this equation are the solutions of the system

$$
\left\{\begin{array} { l } 
{ | x - 3 | > 0 , } \\
{ | x - 3 | \neq 1 , } \\
{ \frac { x + 1 } { 4 } = \frac { x - 2 } { 3 } }
\end{array} \text { , i.e. } \left\{\begin{array}{l}
x \neq 3 \\
x \neq 2, x \neq 4 \\
3 x+3=4 x-8
\end{array}\right.\right.
$$

and, perhaps, the solutions of the equation $|x-3|=1$. The equation of the system has the root $x=11$, and the condition $|x-3|=1$ is satisfied by $x=2$ and $x=4$, which are also solutions of the system, since at these values the functions $\frac{x+1}{4}$ and $\frac{x-2}{3}$ are defined. Thus, $x_{1}=2, x_{2}=4$, $x_{3}=11$.

Answer: $x_{1}=2, x_{2}=4, x_{3}=11$.