in short, (h o g)(a) is just h( g(a) ).
so what we can do is simply get g(a) first and then plug that in h(x).
[tex]\bf \begin{cases}
g(x)&=2x\\
h(x)&=x^2+4\\
(h\circ g)(a)&=h(~~g(a)~~)
\end{cases}
\\\\\\
g(a)=2(a)\implies g(a)=2a
\\\\\\
h(~~g(a)~~)\implies h(~~2a~~)=(2a)^2+4
\\\\\\
h(~~2a~~)=(2^2a^2)+4\implies h(~~2a~~)=4a^2+4[/tex]
