Domain of $f\left(x\right)$ is $\left[1,3\right]$

$1\le {\log }_2\left(x^2+3x-2\right)\le 3$

$\Rightarrow 2\le x^2+3x-2\le 8$

It is the same as solving two inequalities.

Case $1:$ $x^2+3x-2\ge 2$

or, $x^2+3x-4\ge 0$

or, $\left(x+4\right)\left(x-1\right)\ge 0$

$\therefore x\in (-\infty ,-4]\cup [1,\infty )$

Case $2:$ $x^2+3x-2\le 8$

or, $x^2+3x-10\le 0$

or, $\left(x+5\right)\left(x-2\right)\le 0$

$\therefore x\in \left[-5,2\right]$

Considering both Case $1$and Case $2$ we have,$x\in \left[-5,-4\right]\cup \left[1,2\right]$