1: The set $\{x:f(x)=f^{-1}(x)\}=\{0,-1\}$ where $f(x)=(x+1{)}^2-1,x\ge -1$.
2: $f$ is a bijection.

For $f(x)=(x+1{)}^2-1$, $x\ge -1$.$f(x)=f^{-1}(x)$ is equivalent to solving $f(x)=x$ (since if it's increasing, they meet on $y=x$).$(x+1{)}^2-1=x⟹x^2+2x+1-1=x⟹x^2+x=0⟹x(x+1)=0$.$x=0,-1$. Statement-1 is true.Statement-2 says $f$ is a bijection. For $f:[-1,\infty )\to [-1,\infty )$, it is a bijection. However, usually, a reason for the intersection being $\{0,-1\}$ isn't just that it's a bijection, but that it's increasing. But technically, $f$ is not a bijection unless the co-domain is specified.The key says Statement-2 is false (likely because co-domain isn't specified).