Let $f\left(x\right)=2+\left|x\right|-\left|x-1\right|+\left|x+1\right|, x\in R$.

Consider

$\left(S1\right):f^{\text{'}}\left(-\frac{3}{2}\right)+f^{\text{'}}\left(-\frac{1}{2}\right)+f^{\text{'}}\left(\frac{1}{2}\right)+f^{\text{'}}\left(\frac{3}{2}\right)=2$

$\left(S2\right): {\int }_{-2}^2 f\left(x\right)dx=12$

Then,

Given $f\left(x\right)=2+\left|x\right|-\left|x-1\right|+\left|x+1\right|, x\in R$ $\Rightarrow f\left(x\right)=\left\{\begin{array}{ccc}-x& ,& x<-1\\ x+2& ,& -1\le x<0\\ 3x+2& ,& 0\le x<1\\ x+4& ,& x\ge 1\end{array}\right.$

$\Rightarrow f^{\text{'}}\left(x\right)=\left\{\begin{array}{ccc}-1& ,& x<-1\\ 1& ,& -1<x<0\\ 3& ,& 0<x<1\\ 1& ,& x>1\end{array}\right.$

$\therefore f^{\text{'}}\left(-\frac{3}{2}\right)+f^{\text{'}}\left(-\frac{1}{2}\right)+f^{\text{'}}\left(\frac{1}{2}\right)+f^{\text{'}}\left(\frac{3}{2}\right)$

$=-1+1+3+1=4$

i.e. $\left(S1\right)$ is incorrect.

Now  ${\int }_{-2}^{2}f\left(x\right)dx={\int }_{-2}^{-1}f\left(x\right)dx+{\int }_{-1}^{0}f\left(x\right)dx+{\int }_0^{1}f\left(x\right)dx+{\int }_1^{2}f\left(x\right)dx$

$={\left[-\frac{x^2}{2}\right]}_{-2}^{-1}+{\left[\frac{{\left(x+2\right)}^2}{2}\right]}_{-1}^0+{\left[\frac{{\left(3x+2\right)}^2}{6}\right]}_0^1+{\left[\frac{{\left(x+4\right)}^2}{2}\right]}_1^2$

$\left(-\frac{1}{2}+2\right)+\left(2-\frac{1}{2}\right)+\left(\frac{25}{6}-\frac{4}{6}\right)+\left(18-\frac{25}{2}\right)$

$=\frac{3}{2}+\frac{3}{2}+\frac{7}{2}+\frac{11}{2}=12$

i.e. $\left(S2\right)$ is correct.