(7) Given the set $A=\left\{(x, y) \left\lvert\,(x-1)^{2}+(y-2)^{2} \leqslant \frac{4}{5}\right.\right\}, B=$ $\{(x, y)|| x-1|+2| y-2 \mid \leqslant a\}$, and $A \subseteq B$, then the range of the real number $a$ is $\qquad$.
(7) $a \geqslant 2$ Hint: Without loss of generality, let $x-1=x_{1}, y-2=y_{1}$, then
$$
\begin{array}{c}
A=\left\{\left(x_{1}, y_{1}\right) \left\lvert\, x_{1}^{2}+y_{1}^{2} \leqslant \frac{4}{5}\right.\right\}, \\
B=\left\{\left(x_{1}, y_{1}\right)|| x_{1}|+2| y_{1} \mid \leqslant a\right\} .
\end{array}
$$

Let $x_{1}=t \cos \theta, y_{1}=t \sin \theta, 0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant t \leqslant \frac{2}{\sqrt{5}}$, then
$$
\left|x_{1}\right|+2\left|y_{1}\right|=t|\cos \theta|+2 t|\sin \theta| \leqslant \sqrt{5} t \leqslant 2 .
$$

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