Solution: (a) Let $\triangle A B C$ be a right-angled triangle, with its vertices having integer coordinates, and the two legs lying on the sides of these square grids. Let $\angle A=90^{\circ}, A B=m, A C=n$. Consider the rectangle $A B C D$ as shown in Figure 1.
For a polygon
$P$, let $S_{1}(P)$ be the total
area of all the black parts
inside $P$, and $S_{2}(P)$ be the total area of all the white parts.

When $m$ and $n$ are both even or both odd, the black parts of $A B C D$ are symmetric about the midpoint of the hypotenuse $B C$. Therefore,
$$\begin{array}{l} S_{1}(A B C)=S_{1}(B C D), \\ S_{2}(A B C)=S_{2}(B C D) . \end{array}$$

Thus,
$$\begin{aligned} f(m, n) & =\mid S_{\mathrm{i}}(A B C)-S_{2}(A B C)! \\ & \left.=-\frac{1}{2} \right\rvert\, S_{1}(A B C D)-S_{2}(A B C D) ; \end{aligned}$$
When $m$ and $n$ are both odd, $f(m, n)=\frac{1}{2}$.
(b) If $m$ and $n$ are both even or both odd, the conclusion follows from (a). Hence, we can assume $m$ is odd and $n$ is even. As shown in Figure 2, consider the point $L$ on $A B$ such that $A L=m-1$.
Since $m-1$ is even, we have $f(m-1, n)=0$, i.e.,
$$\begin{array}{l} S_{1}(A L C)=S_{2}(A L C) \text {. Therefore, } \\ f(m, n)=\left|S_{1}(A B C)-S_{2}(A B C)\right| \\ =\left|S_{1}(L B C)-S_{2}(L B C)\right| \\ \leqslant \text { the area of } \triangle L B C =\frac{n}{2} \leqslant \frac{1}{2} \max \{m, n\} \text {. } \\ \end{array}$$
(c) We calculate the value of $f(2 k+1,2 k)$. As in (b), consider the point $L$ on $A B$ such that $A L=2 k$. Since $f(2 k, 2 k)=0$ and $S_{1}(A L C)=S_{2}(A L C)$, we have
$$f(2 k+1,2 k)=\left|S_{1}(L B C)-S_{2}(L B C)\right| \cdot \triangle L B C$$

The area of $\triangle L B C$ is $k$. Without loss of generality, we can assume the diagonal $L C$ lies entirely in the black squares
(see Figure 3). Thus, the white part of $\triangle L B C$ consists of several triangles:
$\triangle B L N_{2 k}$,
$$\begin{array}{l} \triangle M_{2 k-1} I_{2 k-1} \mid N_{2 k-1}, \cdots, \\ \triangle M_{1} L_{1} N_{1} \text {, each of which is } \end{array}$$

similar to $\triangle B A C$. Their total area is
Figure 2
$$\begin{array}{l} =\frac{1}{2} \frac{2 k}{2 k+1}\left(\left(\frac{2 k}{2 k}\right)^{2}+\left(\frac{2 k-1}{2 k}\right)^{2}+\cdots+\left(\frac{1}{2 k}\right)^{2}\right) \\ =\frac{1}{4 k(2 k+1)}\left(1^{2}+2^{2}+\cdots+(2 k)^{2}\right) \\ =\frac{4 k+1}{12} \end{array}$$

Therefore, the total area of the black part is
$$S_{1}(L B C)=k-\frac{1}{12}(4 k+1)=\frac{1}{12}(8 k-1) .$$

Finally, we get
$$f(2 k+1,2 k)=\frac{2 k-1}{6} .$$

This function can take arbitrarily large values.