Xét hàm số $f\left( x \right)={{x}^{3}}-3x+m$ trên đoạn $\left[ 0;\,2 \right]$
Ta có ${f}'\left( x \right)=3{{x}^{2}}-3=0$$\Rightarrow x=1.$
Lại có $f\left( 0 \right)=m;\,\,f\left( 1 \right)=m-2;\,\,$$f\left( 2 \right)=m+2$
Do đó $f\left( x \right)\in \left[ m-2;\,\,m+2 \right]$
Nếu $m-2\ge 0$$\Rightarrow \underset{\left[ 0;\,2 \right]}{\mathop{\text{Max}}}\,\left| f\left( x \right) \right|=m+2$$=3m=1\,\,\left( L \right)$
Nếu $m-2<0\Rightarrow \left[ \begin{align} & \underset{\left[ 0;\,2 \right]}{\mathop{\text{Max}}}\,\left| f\left( x \right) \right|=m+2 \\ & \underset{\left[ 0;\,2 \right]}{\mathop{\text{Max}}}\,\left| f\left( x \right) \right|=2-m \\ \end{align} \right.$
+) TH1: $\underset{\left[ 0;\,2 \right]}{\mathop{\text{Max}}}\,\left| f\left( x \right) \right|=m+2=3$
$\Leftrightarrow m=1\Rightarrow 2-m=1 < 3\,\left( tm \right)$
+) TH2: $\underset{\left[ 0;\,2 \right]}{\mathop{\text{Max}}}\,\left| f\left( x \right) \right|=2-m=3$
$\Leftrightarrow m=-1\Rightarrow m+2=1 < 3\,\left( tm \right)$
Vậy $m=1,\,\,m=-1$ thoả mãn yêu cầu bài toán.
Vậy $S$ có 2 phần tử.