Solution set!

$\begin{aligned} \left(\dfrac 13\right)^{\frac {|x+2|}{2-|x|}} & > 9 \\ 3^{-\frac {|x+2|}{2-|x|}} & > 9 \\ 3^{\frac {|x+2|}{|x|-2}} & > 3^2 \\ \implies \frac {|x+2|}{|x|-2} & > 2 \end{aligned}$

Let $\dfrac {|x+2|}{|x|-2} = \begin{cases} \dfrac {-x-2}{-x-2} = 1 \color{#D61F06} < 2 & \text{for }x < -2 \\ \color{#D61F06} \dfrac 00 \implies \text{undefined} & \text{for }x=-2 \\ \dfrac {x+2}{-x-2} = -1 \color{#D61F06} < 2 & \text{for }-2 < x \le 0 \\ \dfrac {x+2}{x-2} & \text{for } x > 0 \end{cases}$

For $0<x<2 \implies -\infty < \dfrac {x+2}{x-2} < -1 \color{#D61F06} < 2$

For $x > 2$ ,

$\begin{aligned} \frac {x+2}{x-2} & > 2 \\ x+2 & > 2x - 4 \\ \implies x & < 6 \end{aligned}$

Therefore, $\left(\dfrac 13\right)^{\frac {|x+2|}{2-|x|}} > 9$ when $x \in (2,6)$ . $\implies a+b = 2+6 = \boxed{8}$ .