$(a):$ Taking $x=0,$ we find $f(f(y))=y+f(0)$ and this means that $f$ is one-to-one and onto. So, $f(f(0))=f(0)$ implies $f(0)=0.$ Hence $f(f(y))=y$ for all $y \in \mathbb{R}.$ Taking $y=f(z)$ in the original equation, we get $f(x+z)=f(x)+f(z)$ and this is the well-known Cauchy equation, and the only solution is $f(x)=cx$ when $f$ is strictly monotone. Substituting, we get $c^2=1,$ so all solutions are $f(x)=x$ and $f(x)=-x.$ Clearly these are strictly monotone. $(b):$ Taking $x=0,$ we get $f(f(y))=y^n+f(0)$ and this means that $f(x)=f(y) \: \Longrightarrow \: x^n=y^n$ So, $f(f(0))=f(0)$ implies $[f(0)]^n=0^n$ and therefore $f(0)=0.$ Therefore $f(f(y))=y^n$ for all $y \in \mathbb{R}$ Taking $y=f(z)$ in the original equation, we get $f(x+z^n)=f(x)+[f(z)]^n$ and by taking $y=f(z)$ in the equation $f(f(y))=y^n,$ we get $f(z^n)=[f(z)]^n$ for all $z \in \mathbb{R}.$ So, $f(x+z^n)=f(x)+[f(z)]^n=f(x)+f(z^n)$ for all $x, \; z \in \mathbb{R}$ So, we can prove that for all $x \in \mathbb{R}^+, \: f(x)=cx$ using the same idea in the part $(a).$ (Note that $n$ can be even.) Therefore, $f(x+z^n)=f(x)+c\cdot z^n, \: \forall x \in \mathbb{R}$ and $\forall z \in \mathbb{R}^+$ Now, we can easily prove that $f(x)=cx, \: \forall x \in \mathbb{R}$ By substituting, we get $c^2y=y^n$ which can only be true for all $y \in \mathbb{R}$ when $n=1.$

Taking $x=0,$ we find $f(f(y))=y+f(0)$ and this means that $f$ is one-to-one and onto. How could you say $f$ is one to one and onto?

To prove it's one-to-one : Assume that $f(a)=f(b)$ for some $a$ and $b$. But then $f(f(a))=f(f(b))$ so $a+f(0)=b+f(0)$ i.e $a=b$ so $f$ is one-to-one To prove it's onto: Notice that plugging $y=x-f(0)$: $$f(f(x-f(0)))=x-f(0)+f(0)=x$$ so $f$ is onto because for every $x$ we can choose $a=f(x-f(0))$ so $f(a)=x$. Hope this helped you

thanks for the help