5. Given a positive geometric sequence $\left\{a_{n}\right\}$ satisfies $a_{7}=a_{6} + 2 a_{5}$. If there exist two terms $a_{m} 、 a_{n}$ such that $\sqrt{a_{m} a_{n}}=4 a_{1}$, then the minimum value of $\frac{1}{m}+\frac{4}{n}$ is $\qquad$
5. $\frac{3}{2}$.

From $a_{7}=a_{6}+2 a_{5}$, we know $q^{2}=q+2$.
Also, since $q>0$, then $q=2$.
Since $\sqrt{a_{m} a_{n}}=4 a_{1}$, we have
$16 a_{1}^{2}=a_{m} a_{n}=a_{1} q^{m-1} a_{1} q^{n-1}$.
Thus, $2^{m+n-2}=2^{4} \Rightarrow m+n=6$.
Therefore, $\frac{1}{m}+\frac{4}{n}=\frac{1}{6}\left(\frac{1}{m}+\frac{4}{n}\right)(m+n)$
$=\frac{1}{6}\left(5+\frac{n}{m}+\frac{4 m}{n}\right) \geqslant \frac{1}{6} \times 9=\frac{3}{2}$,
with equality holding if and only if $m=2, n=4$.
Thus, the minimum value of $\frac{1}{m}+\frac{4}{n}$ is $\frac{3}{2}$.