(5) Given a positive geometric sequence $\left\{a_{n}\right\}$ satisfies: $a_{7}=a_{6}+2 a_{5}$, if there exist two terms $a_{m}$ and $a_{n}$ such that $\sqrt{a_{m} \cdot a_{n}}=4 a_{1}$, then the minimum value of $\frac{1}{m}+\frac{4}{n}$ is $\qquad$.
(5) $\frac{3}{2}$ Hint: From $a_{7}=a_{6}+2 a_{5}$, we get $q^{2}=q+2$. Since $q>0$, we have $q=$
2. Since $\sqrt{a_{m} a_{n}}=4 a_{1}$, we have
$$
16 a_{1}^{2}=a_{m} a_{n}=a_{1} q^{m-1} \cdot a_{1} q^{n-1} .
$$

Therefore, $16=2^{m+n-2}=2^{4}$. Thus, $m+n=6$. So
$$
\begin{aligned}
\frac{1}{m}+\frac{4}{n} & =\frac{1}{6}\left(\frac{1}{m}+\frac{4}{n}\right)(m+n) \\
& =\frac{1}{6}\left(5+\frac{n}{m}+\frac{4 m}{n}\right) \\
& \geqslant \frac{1}{6} \times 9=\frac{3}{2},
\end{aligned}
$$

with equality holding if and only if $m=2, n=4$. Therefore, the minimum value of $\frac{1}{m}+\frac{4}{n}$ is $\frac{3}{2}$.