II. (25 points) Given a two-digit number, whose tens and units digits are $p$ and $q$ respectively, the quadratic function $y=x^{2}+q x+p$ intersects the $x$-axis at two distinct points $A$ and $B$, with the vertex at $C$, and $S_{\triangle A B C} \leqslant 1$.
(1) Find the range of $q^{2}-4 p$;
(2) Find all such two-digit numbers $pq$.
(1) Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)\left(x_{1} \neq x_{2}\right)$, then $x_{1}, x_{2}$ are the two distinct real roots of the equation $x^{2}+q x+p=0$. Therefore,
$$
x_{1}+x_{2}=-q, x_{1} x_{2}=p, q^{2}-4 p>0 \text{. }
$$

Also, $y_{c}=\frac{4 p-q^{2}}{4}\left(y_{c}\right.$ represents the y-coordinate of point $C)$, so,
$S_{\triangle A B C}=\frac{1}{2}\left|x_{1}-x_{2}\right| \cdot\left|y_{c}\right|$
$$
=\frac{1}{2} \cdot \sqrt{q^{2}-4 p} \cdot\left|\frac{4 p-q^{2}}{4}\right| \leqslant 1 \text{. }
$$

Thus, $\left(q^{2}-4 p\right)^{3} \leqslant 64, q^{2}-4 p \leqslant 4$.
Hence $0<q^{2}-4 p \leqslant 4$.
(2) From (1), we know that $q^{2}-4 p=1,2,3,4$. Since $q^{2}$ leaves a remainder of 0 or 1 when divided by 4, $q^{2}-4 p$ also leaves a remainder of 0 or 1 when divided by 4. Therefore, $q^{2}-4 p=1$ or 4. The integer solutions that satisfy the conditions in these two equations are $\left\{\begin{array}{l}p=2, \\ q=3 ;\end{array}\left\{\begin{array}{l}p=6, \\ q=5 ;\end{array}\left\{\begin{array}{l}p=3, \\ q=4 ;\end{array}\left\{\begin{array}{l}p=8, \\ q=6 .\end{array}\right.\right.\right.\right.$
Thus, the two-digit number $\overline{p q}$ is: $23,65,34,86$.