5. Given positive numbers $a$ and $b$ satisfy $a+3b=7$. Then the minimum value of $\frac{1}{1+a}$ $+\frac{4}{2+b}$ is $\qquad$ .
5. $\frac{13+4 \sqrt{3}}{14}$.

Notice,
$$
\begin{array}{l}
\frac{1}{1+a}+\frac{4}{2+b}=\frac{1}{1+a}+\frac{12}{6+3 b} \\
\geqslant \frac{(1+\sqrt{12})^{2}}{1+a+6+3 b}=\frac{13+4 \sqrt{3}}{14} .
\end{array}
$$

Upon verification, the equality holds in the above expression.