Range of $x$

$\left(\frac{1}{3}\right)^{-2}=9,$ and so the expression in the exponent must be less than $-2:$

$\frac{\vert x \vert +1}{x-2} < -2$

Suppose that $x>2.$ Then,

$\begin{aligned} x+1 &< -2(x-2) \\ 3x &< 3 \\ x &< 1 \quad {\color{#D61F06}\text{A contradiction!}} \end{aligned}$

Now suppose that $x<2.$ Then,

$|x|+1 > -2(x-2)$

If $x>0:$

$\begin{aligned} x+1 &> -2(x-2) \\ 3x &> 3 \\ x &> 1 \\ \end{aligned}$

If $x<0:$

$\begin{aligned} -x+1 &> -2(x-2) \\ x &> 3 \quad {\color{#D61F06}\text{A contradiction!}} \end{aligned}$

Therefore, the interval that satisfies the inequality is: $x \in (1,2).$