$$\begin{gathered} Let S=\left\{x\in \left[-6, 3\right]-\left\{-2, 2\right\}:\frac{\left|x+3\right|-1}{\left|x\right|-2}\ge 0\right\} \\ and T=\left\{x\in Z:x^2-7\left|x\right|+9\le 0\right\}. Then the \\ number of elements is S\cap T is \end{gathered}$$

$$\begin{gathered} case1: \\ \left|x\right|-2>0 and\left|x+3\right|-1\ge 0 \\ \Rightarrow \left|x\right|>2 \Rightarrow x>2 or x<-2 \\ and\left|x+3\right|\ge 1\Rightarrow x+3\ge 1 or x+3\le -1\Rightarrow x\ge -2 or x\le -4 \\ \Rightarrow x\in \left[-6,-4\right]\cup (2,3] \\ case2: \\ \left|x\right|-2<0 and\left|x+3\right|-1\le 0 \\ \Rightarrow \left|x\right|<2\Rightarrow -2<x<2 and\left|x+3\right|\le 1\Rightarrow -1\le x+3\le 1\Rightarrow -4\le x\le -2 \\ \therefore No common solution exists \\ \Rightarrow S=\left\{x\in \left[-6,-4\right]\cup (2,3]\right\} \end{gathered}$$

$$\begin{gathered} T=\left\{x\in Z:x^2-7\left|x\right|+9\le 0\right\} \\ {x}^2-7x+9\le 0 for x\ge 0\Rightarrow x\in \left\{2,3,4,5\right\} \\ {x}^2+7x+9\le 0 for x\le 0\Rightarrow x\in \left\{-5,-4,-3,-2\right\} \\ T=\left\{\pm 2,\pm 3,\pm 4,\pm 5\right\} \\ \mathrm{Thus} \mathrm{S}\cap \mathrm{T}=\left\{-5,-4,3\right\} \\ \mathrm{and} \mathrm{S}\cap \mathrm{T} \mathrm{contains} \mathrm{exactly} 3 \mathrm{elements} \end{gathered}$$