We have,
$$
\begin{aligned}
& f_1(x)=|x|=\left\{\begin{array}{cc}
-x, & x < 0 \\
x, & x \geq 0
\end{array} \Rightarrow f_1^{\prime}(x)= \begin{cases}-1, & x < 0 \\
1, & x \geq 0\end{cases} \right. \\
& \therefore f_1^{\prime}(1)=1 \\
& f_2(x)=\left\{\begin{array}{cc}
1+\sin (x-1), & x \leq 1 \\
x, & x \geq 1
\end{array}\right. \\
& \Rightarrow f_2^{\prime}(x)=\left\{\begin{array}{cc}
\cos (x-1), & x \leq 1 \\
1, & x \geq 1
\end{array}\right. \\
& \therefore f_2^{\prime}(1)=1 \\
& f_3(x)=\left\{\begin{array}{cc}
x^2+7 x-7, & x \leq 1 \\
\frac{3 x-1}{2}, & x \geq 1
\end{array}\right. \\
& \Rightarrow f_3^{\prime}(x)=\left\{\begin{array}{cc}
2 x+7, & x \leq 1 \\
\frac{3}{2}, & x \geq 1
\end{array}\right. \\
&
\end{aligned}
$$
Since, LHD (at $x=1)=2 \times 1+7=9$
$$
\begin{aligned}
& \text { and RHD (at } x=1)=\frac{3}{2} \\
\therefore \quad \text { LHD } & \neq \text { RHD }
\end{aligned}
$$
Hence, $f_3(x)$ is not differentiable at $x=1$
$$
\begin{aligned}
& f_4(x)=\left\{\begin{array}{cc}
|x-1|+|x-2|, & x \leq 1 \\
1+x-x^3, & x \geq 1
\end{array}\right. \\
& =\left\{\begin{array}{cc}
-(x-1)-(x-2, & x \leq 1 \\
1+x-x^3, & x \geq 1
\end{array}=\left\{\begin{array}{cc}
3-2 x, & x \leq 1 \\
1+x-x^3, & x \geq 1
\end{array}\right.\right. \\
& \Rightarrow \quad f_4^{\prime}(x)=\left\{\begin{array}{cc}
-2, & x \leq 1 \\
1-3 x^2, & x \geq 1
\end{array}\right. \\
& \therefore \quad f_4^{\prime}(1)=-2 \\
&
\end{aligned}
$$