Three, (18 points) Given that the linear function $y=A x+B$ and the inverse proportion function $y=\frac{k}{x}$ intersect at points $M(2,3)$ and $N(-4, m)$.
(1) Find the expressions for the linear function $y=A x+B$ and the inverse proportion function $y=\frac{k}{x}$;
(2) Find the area of $\triangle M O N$.
Three, (1) Given that $M(2,3)$ is in the first quadrant, we know that $N(-4, m)$ is in the second or third quadrant. Draw the schematic diagram of the linear function $y=A x+B$ $(A>0)$ and the inverse proportion function $y=\frac{k}{x}(k>0)$ (as shown in Figure 10).

Since $M(2,3)$ and $N(-4, m)$ are both on the graph of $y=\frac{k}{x}$, we have
\[
\left\{\begin{array}{l}
3=\frac{k}{2}, \\
m=\frac{k}{-4}.
\end{array}\right.
\]
Solving these, we get
\[
\left\{\begin{array}{l}
k=6, \\
m=-\frac{3}{2}.
\end{array}\right.
\]
Thus, we have $N\left(-4,-\frac{3}{2}\right)$.
Since $M(2,3)$ and $N\left(-4,-\frac{3}{2}\right)$ are on the graph of $y=A x+B$, we have
\[
\left\{\begin{array}{l}
3=2 A+B, \\
-\frac{3}{2}=-4 A+B.
\end{array}\right.
\]
Solving these, we get
\[
\left\{\begin{array}{l}
A=\frac{3}{4}, \\
B=\frac{3}{2}.
\end{array}\right.
\]
Therefore, the required equations are $y=\frac{3}{4} x+\frac{3}{2}$ and $y=\frac{6}{x}$.
(2) Let the line $M N$ intersect the $x$-axis and $y$-axis at points $P$ and $Q$ respectively.
Substituting $y=0$ and $x=0$ into $y=\frac{3}{4} x+\frac{3}{2}$, we get $x=-2$ and $y=\frac{3}{2}$, so $P(-2,0)$ and $Q\left(0, \frac{3}{2}\right)$.
Thus, $S_{\triangle M O N}=S_{\triangle M O Q}+S_{\triangle Q O P}+S_{\triangle P O N}$
\[
\begin{aligned}
= & \left.\frac{1}{2} \times \frac{3}{2} \times 2+\frac{1}{2} \times 1-2 \right\rvert\, \times \frac{3}{2}+ \\
& \left.\frac{1}{2} \times 1-2|\times|-\frac{3}{2} \right\rvert\,=\frac{9}{2} .
\end{aligned}
\]