II. (16 points) Given that $m$ and $n$ are positive integers. If the two real roots of the equation $4 x^{2}-2 m x+n=0$ are both greater than 1 and less than 2, find the values of $m$ and $n$.
Let $f(x)=4 x^{2}-2 m x+n$, then the graph of $y=f(x)$ is a parabola opening upwards, with the axis of symmetry at $x=\frac{m}{4}$.
$$
\begin{array}{l}
10, \\
f(2)=16-4 m+n>0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
m^{2} \geqslant 4 n, \\
4+n>2 m,
\end{array}\right. \\
\hline
\end{array}
$$

From (1), we know $m=5,6,7$.
When $m=5$, from (2) we get $n \leqslant \frac{25}{4}$, so $n \leqslant 6$.
Also, from (3) we get $n>6$. This is a contradiction.
When $m=6$, from (2) we get $n \leqslant 9$.
Also, from (3) and (4) we get $n>8$, so $n=9$.
When $m=7$, from (2) we get $n \leqslant 12$.
Also, from (4) we get $n>12$, which is a contradiction.
In summary, $m=6, n=9$.