(5) The graph of the function $y=a^{x+3}-2(a>0$ and $a \neq 1)$ always passes through a fixed point $A$. If point $A$ lies on the line $\frac{x}{m}+\frac{y}{n}+1=0$, and $m, n>0$, then the minimum value of $3 m+n$ is $\qquad$
(5) 16 Hint: Since the graph of the function $y=a^{x+3}-2(a>0$ and $a \neq 1)$ always passes through the fixed point $(-3,-1)$, the coordinates of point $A$ are $(-3,-1)$, thus $\frac{-3}{m}+\frac{-1}{n}+1=0$, which means $1=\frac{3}{m}+\frac{1}{n}$. Therefore,
$$
3 m+n=(3 m+n) \cdot\left(\frac{3}{m}+\frac{1}{n}\right)=10+\frac{3 n}{m}+\frac{3 m}{n} \geqslant 16 .
$$

Equality holds if and only if $m=n=4$. Therefore, the minimum value of $3 m+n$ is 16.