7. Given the sets
$$
\begin{array}{l}
A=\left\{(x, y) \left\lvert\,(x-1)^{2}+(y-2)^{2} \leqslant \frac{4}{5}\right.\right\}, \\
B=\{(x, y)|| x-1|+2| y-2 \mid \leqslant a\},
\end{array}
$$

and $A \subseteq B$. Then the range of the real number $a$ is
$$
\text { II.7. } a \geqslant 2 \text {. }
$$

Assume $x-1=x_{1}, y-2=y_{1}$. Then
$$
\begin{array}{l}
A=\left\{\left(x_{1}, y_{1}\right) \left\lvert\, x_{1}^{2}+y_{1}^{2} \leqslant \frac{4}{5}\right.\right\}, \\
B=\left\{\left(x_{1}, y_{1}\right)|| x_{1}|+2| y_{1} \mid \leqslant a\right\} . \\
\text { Let } x_{1}=t \cos \theta, y_{1}=t \sin \theta, \text { where, } \\
0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant t \leqslant \frac{2}{\sqrt{5}} .
\end{array}
$$
$$
\begin{array}{l}
\text { Then }\left|x_{1}\right|+2\left|y_{1}\right|=t|\cos \theta|+2 t|\sin \theta| \\
\leqslant \sqrt{5} t \leqslant 2 .
\end{array}
$$

From $A \subseteq B$, we know $a \geqslant 2$.