7. Given positive numbers $a, b$ satisfy $2 a+b=1$, then the maximum value of $4 a^{2}+b^{2}+4 \sqrt{a b}$ is $\qquad$
$$
\begin{array}{l}
(2 a+b)^{2}=4 a^{2}+b^{2}+4 a b=1 \Rightarrow 4 a^{2}+b^{2}+4 \sqrt{a b}=1-4 a b+4 \sqrt{a b}, \\
\text { Also } 1=2 a+b \geqslant 2 \sqrt{2 a b} \Rightarrow \sqrt{a b} \leqslant \frac{1}{2 \sqrt{2}},
\end{array}
$$

Therefore, the maximum value of $4 a^{2}+b^{2}+4 \sqrt{a b}$ is $\frac{1}{2}+\sqrt{2}$, with equality when $a=\frac{1}{4}, b=\frac{1}{2}$.