Answer
$f(x)=\sqrt{1+x}$
And:
$g(x)=\sqrt{x}$
Other answers are also possible.
Work Step by Step
We want: $H(x)=(f\circ g)(x)=f(g(x))$.
We make:
$f(x)=\sqrt{1+x}$
And:
$g(x)=\sqrt{x}$
Then:
$f(g(x))=\sqrt{1+\sqrt{x}}=H(x)$
Other answers are also possible.