$\text{ The function }f\left(x\right)=\left\{\begin{array}{ll}\frac{\pi }{4}+{\tan }^{-1}x,& \left|x\right|\le 1\\ \frac{1}{2}\left(\right|x|-1),& \left|x\right|>1\end{array}\right.\text{ is: }$

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{ll}\frac{\pi }{4}+{\tan }^{-1}x& \left|x\right|\le 1\\ \frac{1}{2}\left(\right|x|-1)& \left|x\right|>1\end{array}\right. \Rightarrow f\left(x\right)=\left\{\begin{array}{l}\frac{1}{2}(x-1) x>1 \\ \frac{\mathrm{\pi }}{4}+{\tan }^{-1}x -1\le x\le 1\\ \frac{1}{2}(-x-1) x<-1\end{array}\right. \\ \Rightarrow f\text{'}\left(x\right)=\left\{\begin{array}{l}\frac{1}{2} x>1\\ \frac{1}{1+x^2} -1<x<1\\ -\frac{1}{2} x<-1\end{array}\right. \end{gathered}$$

$$\begin{gathered} it is clear that we need to check the continuity at x=-1,1 \\ now f(-1)=f(-1^{+})=\frac{\mathrm{\pi }}{4}+{\tan }^{-1}(-1)=\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}=0 \\ f(-1^{-})=\frac{1}{2}(1-1)=0 \\ \Rightarrow f(-1^{-})=f(-1)=f(-1^{+}) \\ \Rightarrow f is continuous at x=-1 and \\ f\text{'}(-1^{-})=-\frac{1}{2}, f\text{'}(-1^{+})=\frac{1}{2} \Rightarrow f\text{'}(-1^{-})\ne f\text{'}(-1^{+}) \\ \Rightarrow f is not differentiable at x=-1 \\ f\left(1\right)=f\left(1^{-}\right)=\frac{\mathrm{\pi }}{4}+{\tan }^{-1}\left(1\right)=\frac{\mathrm{\pi }}{2} and f\left(1^{+}\right)=0 \\ \Rightarrow f\left(1\right)=f\left(1^{-}\right)\ne f\left(1^{+}\right) \\ \Rightarrow f is discontinuous at x=1 \Rightarrow f is also non differentiable at x=1 \end{gathered}$$

$\text{ So continuous on }\mathrm{R}-\left\{1\right\}\text{ and differentiable }\mathrm{R}-\{-1,1\}$