II. (25 points) As shown in Figure 4, in the Cartesian coordinate system, the vertices of rectangle $O A B C$ are $A$ and $C$ on the $x$-axis and $y$-axis, respectively, and vertex $B$ is in the first quadrant. Points $E$ and $F$ are on $O A$ and $A B$, respectively. When $\triangle A E F$ is folded along $E F$, point $A$ exactly lands on point $D$ on line segment $B C$. Every line passing through the vertex $P$ of the parabola $\Gamma$: $y=a x^{2}-14 a x+49 a+4(a \neq 0)$ always bisects the perimeter of rectangle $O A B C$. If point $P$ lies on line segment $D E$, the length of $A F$ is an integer, and the parabola $\Gamma$ intersects line segment $E F$ at two distinct points, find the range of the real number $a$.
II. Since $y=a x^{2}-14 a x+49 a+4=a(x-7)^{2}+4$, therefore, $P(7,4)$.
As shown in Figure 9, we easily get
$$
\begin{array}{l}
A(14,0), \\
B(14,8), \\
C(0,8) .
\end{array}
$$

Let $A F^{\prime}=x(x$ be an integer), then
$$
\begin{array}{l}
D F=x, B F=8-x, \\
E D=E A=14-O E \\
=14-B D \\
=14-\sqrt{x^{2}-(8-x)^{2}}=14-4 \sqrt{x-4} .
\end{array}
$$

Draw $E G \perp B C$ at point $G$, then
$\triangle E G D \backsim \triangle D B F$.
Thus, $\frac{D F}{E D}=\frac{B D}{E G}$, that is
$$
\frac{x}{14-4 \sqrt{x-4}}=\frac{4 \sqrt{x-4}}{8} \text {. }
$$

Simplifying and rearranging gives
$$
3(x-4)-7 \sqrt{x-4}+4=0 \text {. }
$$

Solving yields $x_{1}=5, x_{2}=\frac{52}{9}$ (discard).
Hence $F(14,5), E(4,0)$.
Since the parabola $\Gamma$ intersects the line segment $E F$ at two distinct points, it is sufficient to ensure that the left intersection point of the parabola $\Gamma$ with the $x$-axis is at point $E$ or to its right. Therefore, we have
$$
a \cdot 4^{2}-14 a \cdot 4+49 a+4 \leqslant 0 .
$$

Solving gives $a \leqslant-\frac{4}{9}$.