Answer
$ f\circ g(x)=2^{x+1},\qquad(-\infty,+\infty$)
$ g\circ f(x)=2^{x}+1,\qquad(-\infty,+\infty$)
Work Step by Step
$f\circ g(x)=f(g(x))=2^{g(x)}=2^{x+1},$
no restriction,defined for all reals
domain=$\mathbb{R}=(-\infty,+\infty$)
$g\circ f(x)=g(f(x)=f(x)+1=2^{x}+1$
no restriction,defined for all reals
domain=$\mathbb{R}=(-\infty,+\infty$)