[tex]\bf \begin{cases}
h(x)=(f\circ g)(x)\\
h(x)=\sqrt{x+5}\\
f(x)=\sqrt{x+2}
\end{cases}\\\\
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(f\circ g)(x)\implies f[\ g(x)\ ]\implies f[\ g(x)\ ]=\sqrt{g(x)+2}\qquad thus
\\\\\\
\sqrt{g(x)+2}=(f\circ g)(x)=h(x)=\sqrt{x+5}\qquad therefore
\\\\\\
\sqrt{g(x)+2}=\sqrt{x+5}\impliedby \textit{squaring both sides}\\\\\\
g(x)+2=x+5\implies g(x)=x+5-2\implies \boxed{g(x)=x+3}[/tex]
