Let f(x)=-1+|x-2| , and g(x)=1-|x| ; then the set of all points...

$$\text{ }$$\begin{aligned} & f \circ g=f(g(x))=f(1-|x|) \\ & =-1+|1-| x|-2| \\ & =-1+|-| x|-1|=-1+|| x|+1| \end{aligned} Let $f \circ g=y$ \begin{aligned} & \therefore y=-1+|| x|+1| \\ & \Rightarrow y= \begin{cases}-1+x+1, & x \geq 0 \\ -1-x+1, & x < 0\end{cases} \\ & \Rightarrow y=\left\{\begin{array}{cc} x, & x \geq 0 \\ -x, & x < 0 \end{array}\right. \end{aligned}$$$LHL at$(x=0)=\lim _{x \rightarrow 0}(-x)=0$RHL at$(x=0)=\lim _{x \rightarrow 0}(x)=0$When$x=0$, then$y=0$Hence, LHL at$(x=0)=$RHL at$(x=0)=$value of$y$at$(x=0)$Hence$y$is continuous at$x=0$. Clearly at all other point$y$ continuous. Therefore, the set of all points where fog is discontinuous is an empty set.