7. Given the function $f(x)$ satisfies:
$$
\begin{array}{l}
f(m+n)=f(m) f(n), f(1)=3 . \\
\text { Then } \sum_{k=1}^{4} \frac{f^{2}(k)+f(2 k)}{f(2 k-1)}=
\end{array}
$$
$$
\text { Notice that the function prototype satisfying } f(m+n)=f(m) f(n), f(1)=3 \text { is the exponential function } y=3^{x}.
$$
$$
\text { Therefore, } \sum_{k=1}^{4} \frac{f^{2}(k)+f(2 k)}{f(2 k-1)}=\sum_{k=1}^{4} \frac{3^{2 k}+3^{2 k}}{3^{2 k-1}}=24 \text {. }
$$