$\text{ Let }f\left(x\right)=|x-1|\text{ . Then }$

$\begin{array}{l}f\left(x\right)=|x-1|=\left\{\begin{array}{cc}-x+1,& x<1\\ x-1,& x\ge 1\end{array}\right.\\ \text{ Consider }f\left(x^2\right)={\left[f\left(x\right)\right]}^2\\ \text{If it is true for all }x,\text{ put }x=2\\ f\left(2^2\right)=\left|4-1\right|=3\\ {\left[f\left(2\right)\right]}^2=1\\ \text{hence the option (1) is not correct}\\ \text{ Consider }f(x+y)=f\left(x\right)+f\left(y\right)\\ \text{ Put }x=2,y=5\text{ we get }\\ f\left(7\right)=6;f\left(2\right)+f\left(5\right)=1+4=5\\ \text{hence, option }\left(2\right)\text{\hspace{0.17em} also not correct}\\ \text{ Consider }f\left(\right|x\left|\right)=\left|f\right(x\left)\right|\\ \text{ Put }x=-5\text{ then }f\left(\right|-5\left|\right)=\text{f}\left(5\right)=4\\ \left|f\right(-5\left)\right|=|-5-1|=6\\ \text{Hence the option (3) is not correct}\end{array}$