[Solution] We only need to solve problem (3) and find all $(x, y)$ that satisfy the equality in (3). For any $(x, y)$, there clearly exist $m^{\prime}, n^{\prime} \in \mathbb{Z}$ such that $\left|x^{\prime}=x-m^{\prime}\right| \leqslant \frac{1}{2},\left|y^{\prime}\right|=$ $\left|y-n^{\prime}\right| \leqslant \frac{1}{2}$ and
$$\overline{f_{a}}(x, y)=\overline{f_{a}}\left(x^{\prime}, y^{\prime}\right)$$

Therefore, we only need to discuss $\overline{f_{a}}(x, y)$ under the condition $|x| \leqslant \frac{1}{2},|y| \leqslant \frac{1}{2}$.
When $a=0$, since $|x| \leqslant \frac{1}{2},|y| \leqslant \frac{1}{2}$, it is obvious that
$$f_{0}(x, y)=x^{2}+y^{2} \leqslant \frac{1}{2}$$

The equality holds if and only if $|x|=\frac{1}{2},|y|=\frac{1}{2}$. When $a=2$,
$$\begin{aligned} f_{2}(x, y) & =\min _{m, 1 \in \mathbb{Z}} f_{2}(x-m, y-n) \\ & =\min _{m, n \in \mathbb{Z}}(x-m)^{2}+2(x-m)(y-n)+(y-n)^{2} \\ & =\min _{m, n \in \mathbb{Z}}(x+y-m-n)^{2} \leqslant \frac{1}{4}, \end{aligned}$$

The equality holds if and only if $|x+y|=\frac{1}{2}$.
We now discuss the case $0<a<2$. It is easy to see that
$$\begin{aligned} f_{a}(x-m, y-n) & =(x-m)^{2}+a(x-m)(y-n)+(y-n)^{2} \\ & =\left[\left(x+\frac{a}{2} y\right)-\left(m+\frac{a}{2} n\right)\right]^{2}+(b y-b n)^{2} \end{aligned}$$

where $b=\frac{1}{2} \sqrt{4-a^{2}}$. Let
$$\left\{\begin{array}{l} u=x+\frac{a}{2} y \\ v=b y \end{array}\right.$$

Let $|x| \leqslant \frac{1}{2},|y| \leqslant \frac{1}{2}$, then $\left|v-\frac{2 b}{a} u\right|=$ $\left|\frac{2 b}{a} x\right| \leqslant \frac{b}{a},|v| \leqslant \frac{b}{2}$, which means all $(u, v)$ form a parallelogram $A B C D$ in the $u v$ plane. After simple calculations, we have
$$A\left(\frac{1}{2}+\frac{a}{4}, \frac{b}{2}\right), B\left(-\frac{1}{2}+\frac{a}{4}, \frac{b}{2}\right), C\left(-\frac{1}{2}-\frac{a}{4},-\frac{b}{2}\right),$$
$D\left(\frac{1}{2}-\frac{a}{4},-\frac{b}{2}\right)$. Since the parallelogram $A B C D$ is centrally symmetric about the origin $O$, we only need to discuss the upper half of the parallelogram $A B E F$, where $E\left(-\frac{1}{2}, 0\right), F\left(-\frac{1}{2}, 0\right)$. Draw a line through the origin $O$ parallel to $B C$ and intersecting the lines $v=$ $\frac{b}{2}$ and $v=b$ at $G$ and $N$, respectively, and extend $N A$ to intersect the $u$-axis at $M$. It is easy to see that $G\left(\frac{a}{4}, \frac{b}{2}\right)$, $N\left(\frac{a}{2}, b\right), M(1,0)$
$$\begin{array}{l} \text { Clearly }\left\{(x, y) ; 0 \leqslant y \leqslant \frac{1}{2},-\frac{1}{2} \leqslant x \leqslant 0\right\} \text { corresponds to } \\ \left\{(u, v) ; 0 \leqslant v \leqslant \frac{b}{2},-\frac{b}{a} \leqslant \frac{2 b}{a} u-v \leqslant 0\right\}, \end{array}$$

which is the parallelogram $G B E O$ and its boundary. Also, $f_{a}(x-m, y-n)$ is equal to the square of the distance from the corresponding point $(u, v)$ to the point $\left(m+\frac{a}{2} n, b n\right)$. Therefore, it is easy to see that
$$\begin{array}{l} f_{a}(x, y)=\left(x+\frac{a}{2} y\right)^{2}+b^{2} y^{2}=u^{2}+v^{2} . \\ \text { Since } E O=G O=\frac{1}{2}, \\ B O^{2}=\left(-\frac{1}{2}+\frac{a}{4}\right)^{2}+\frac{b^{2}}{4}=\frac{1}{4}+\frac{a^{2}}{16}-\frac{a}{4}+\frac{4-a^{2}}{16}=\frac{1}{2}-\frac{a}{4} . \end{array}$$

Thus, we have
$\bar{f}_{a}(x, y) \leqslant \max \left(\frac{1}{4}, \frac{1}{2}-\frac{a}{4}\right)$, for any
$$-\frac{1}{2} \leqslant x \leqslant 0,0 \leqslant y \leqslant \frac{1}{2}$$

If $(x, y)$ satisfies $0 \leqslant x \leqslant \frac{1}{2}, 0 \leqslant y \leqslant \frac{1}{2}$, then the corresponding $(u, v)$ is in the parallelogram $A G O F$ or on its boundary. It is not difficult to prove that $\overline{f_{a}}(x, y)$ is equal to the smallest of the squares of the distances from $(u, v)$ to the points $M, N$, and $O$. Therefore, when
$$\begin{array}{l} 0 \leqslant x \leqslant \frac{1}{2}, 0 \leqslant y \leqslant \frac{1}{2} \text {, } \\ \overline{f_{a}}(x, y) \leqslant R^{2}, \end{array}$$

where $R$ is the circumradius of $\triangle M N O$, and the equality holds if and only if $(u, v)$ is the circumcenter of $\triangle M N O$. After simple calculations, it is easy to see that $R^{2}=\frac{1}{a+2}$, and the circumcenter of $\triangle M N O$ is $\left(\frac{1}{2}, \frac{b}{2+a}\right)$. By
$$\left\{\begin{array}{l} x+\frac{a}{2} y=\frac{1}{2} \\ b y=\frac{b}{2+a} \end{array}\right.$$

we solve $x=y=\frac{1}{2+a}$.
It is clear that $\max \left(\frac{1}{4}, \frac{1}{2}-\frac{a}{4}\right)<\frac{1}{a+2}$, for any $0<a<2$.
In summary, for any $a \in[0,2]$, there exists $c=\frac{1}{a+2}$ such that for any $(x, y)$,
$$\overline{f_{a}}(x, y) \leqslant c \text {. }$$

If $x=x^{\prime}+m, y=y^{\prime}+n$, where $\left|x^{\prime}\right| \leqslant \frac{1}{2},\left|y^{\prime}\right| \leqslant \frac{1}{2}, m, n \in \mathbb{Z}$, then $\overline{f_{a}}(x, y)=c \Leftrightarrow$ when $a=0$, $\left|x^{\prime}\right|=\left|y^{\prime}\right|=\frac{1}{2}$; when $a=2$, $\mid x^{\prime}+$ $y^{\prime} \mid=\frac{1}{2}$; when $0<a<2$, $x^{\prime}=\frac{1}{a+2}, y^{\prime}=\frac{1}{a+2}$ or $x^{\prime}=-\frac{1}{a+2}$, $y^{\prime}=-\frac{1}{a+2}$.