Answer
(a.)$(f\circ g)(x) =\sqrt {x+2}$.
(b.)$(g\circ f)(x) =\sqrt x+2$.
(c.)$(f\circ g)(2) =2$.
Work Step by Step
The given functions are
$f(x)=\sqrt x$ and $g(x)=x+2$.
(a) $(f\circ g)(x) = f(g(x))$
Replace $x$ with $g(x)$ in the function $f(x)$.
$f(g(x))=\sqrt {g(x)}$
Plug value of $g(x)$ in the right hand side.
$f(g(x))=\sqrt {x+2}$
Hence, $(f\circ g)(x) =\sqrt {x+2}$.
(b) $(g\circ f)(x) = g(f(x))$
Replace $x$ with $f(x)$ in the function $g(x)$.
$g(f(x))=f(x)+2$
Plug value of $f(x)$ in the right hand side.
$g(f(x))=\sqrt x+2$.
Hence, $(g\circ f)(x) =\sqrt x+2$.
(c) Replace $x$ with $2$ in the part (a) solution.
$(f\circ g)(2) =\sqrt {2+2}$
$(f\circ g)(2) =\sqrt {4}$
$(f\circ g)(2) =2$.