II. (25 points) As shown in Figure 4, in the Cartesian coordinate system $x O y$, the vertices of the polygon $O A B C D E$ are
$$
\begin{array}{l}
O(0,0), A(0,6), B(4,6), \\
C(4,4), D(6,4), E(6,0) .
\end{array}
$$

It is known that the line $l$ passes through point $M$, intersects sides $O A$ and $D E$, and divides the polygon $O A B C D E$ into two parts of equal area.
(1) If point $M\left(\frac{7}{2}, \frac{5}{2}\right)$, find the function expression of the line $l$;
(2) Prove that if point $M\left(3, \frac{8}{3}\right)$, then there are infinitely many lines $l$ that divide the polygon $O A B C D E$ into two parts of equal area.
II. As shown in Figure 7, extend $BC$ to intersect the $x$-axis at point $F$, connect $OB$ and $AF$ to intersect at point $P$, and connect $CE$ and $DF$ to intersect at point $Q$.

It is easy to see that $P(2,3)$ and $Q(5,2)$ are the centers of rectangles $OFBA$ and $FEDC$, respectively.

Therefore, a line through point $P$ divides rectangle $OFBA$ into two equal areas, and a line through point $Q$ divides rectangle $FEDC$ into two equal areas.
Let $PQ$ intersect $OA$ and $DE$ at points $G$ and $H$, respectively.
Thus, line $PQ$ divides the polygon $OABCDE$ into two equal areas.
Hence, the function expression of line $PQ$ is
$$
y=-\frac{1}{3} x+\frac{11}{3} \text {. }
$$
(1) Clearly, point $M\left(\frac{7}{2}, \frac{5}{2}\right)$ lies on $y=-\frac{1}{3} x+\frac{11}{3}$.

Since $M$ is not the midpoint of segment $GH$, any other line through point $M$ cannot divide the polygon $OABCDE$ into two equal areas.
Therefore, the function expression of line $l$ is $y=-\frac{1}{3} x+\frac{11}{3}$.
(2) Consider another line through point $M\left(3, \frac{8}{3}\right)$ that intersects $OA$ and $DE$ at points $G'$ and $H'$, respectively.
Note that the midpoint of segment $G'H'$ is $M\left(3, \frac{8}{3}\right)$. Hence, $G'H'$ is also a line that satisfies the condition.
By the arbitrariness of $G'H'$, we know that there are infinitely many lines that satisfy the condition.