$$
\text { } \begin{aligned}
& f \circ g=f(g(x))=f(1-|x|) \\
& =-1+|1-| x|-2| \\
& =-1+|-| x|-1|=-1+|| x|+1|
\end{aligned}
$$
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.