$Let f\left(x\right)={(x+1)}^2-1,x\ge -1.$
$\begin{array}{l}Statement1:Theset\{x:f\left(x\right)=f^{-1}\left(x\right)\}=\{0,-1\}.\\ Statement2:fisabijection.\end{array}$

$\begin{array}{l}f\left(x\right)={(x+1)}^2-1,x\ge -1.\\ \Rightarrow f^{\text{'}}\left(x\right)=2(x+1)\ge 0forx\ge -1\\ \Rightarrow f\left(x\right)isone-one\end{array}$

Since co-domain of the given function is not given, hence it can be considered as R, the set of reals and consequently R is not onto.

$\begin{array}{l}Also\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}f\left(x\right)={(x+1)}^2-1,x\ge -1forx\ge -1\\ \Rightarrow R_f=[-1,\infty )\\ Clearlyf\left(x\right)=f^{-1}\left(x\right)atx=0andx=-1.\\ Statement-1istrue\end{array}$