Solutions

a) Yes, any function $g$ of the form $g(x)=x+c$ where $c$ is any real number. This is because addition is commutative, so that

$$\begin{align*} f(g(x))&=f(x+c) \\ &=(x+c)+1 \\ &=(x+1)+c \\ &=g(x+1) \\ &=g(f(x)). \end{align*}$$

b) Suppose $f$ is a constant function defined as $f(x)=c$. Then we have that $f(g(x))=c$, which means that $g(c)=c$. Hence, $g$ is either the same constant function as $f$ or is the identity function.

c) Consider the set of functions $G=\{g(x)=c:c\in\mathbb{R}\}$. Then, for all $g\in G$, we have that

$$f(g(x))=f(c)=c=g(f(x)),$$

which implies that $f(x)=x$ for all $x$.