Three. (50 points) As shown in Figure 3, in the acute triangle $\triangle ABC$, $\angle B > \angle C$, $H$, $I$, and $O$ are the orthocenter, incenter, and circumcenter, respectively. $M$ is the midpoint of $AI$. Draw $AD \perp BC$ at point $D$, such that $\angle DAM = 2 \angle ADM$. Try to express $\angle HIO$ in terms of $\angle B$.

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Three, as shown in Figure 4, construct $I E \perp B C$ at point $E$, $M N \perp B C$ at point $N$, $I F \perp A D$ at point $F$, $I K \perp A B$ at point $K$, and construct the angle bisector of $\angle D A M$ intersecting $M D$ at point $G$.
Figure 4
Let $A M=M I=x, A D=h, I E=r$.
From $\angle D A M=2 \angle A D M$
$\Rightarrow \angle M A G=\angle D A G=\angle A D M$
$\Rightarrow \triangle A M G \backsim \triangle D M A$
$\Rightarrow M A^{2}=M G \cdot M D$.
By the Angle Bisector Theorem, we get
$$
\begin{array}{l}
\frac{A M}{A D}=\frac{M G}{D G} \Rightarrow \frac{M G}{M D}=\frac{x}{x+h} \\
\Rightarrow M G=\frac{x}{x+h} M D \\
\Rightarrow M A^{2}=\frac{x}{x+h} M D^{2} \\
\Rightarrow x^{2}=\frac{x}{x+h} M D^{2} \\
\Rightarrow x^{2}+h x=M D^{2} .
\end{array}
$$

Also, $M D^{2}=M N^{2}+D N^{2}$
$=\left(\frac{A D+I E}{2}\right)^{2}+\left(\frac{I F}{2}\right)^{2}$
$=\frac{(h+r)^{2}}{4}+\frac{A I^{2}-A F^{2}}{4}$
$=\frac{(h+r)^{2}}{4}+\frac{4 x^{2}-(h-r)^{2}}{4}$
$=\frac{4 x^{2}+4 h r}{4}=x^{2}+h r$
$\Rightarrow x^{2}+h x=x^{2}+h r$
$\Rightarrow x=r$.
Thus, $\sin \angle K A I=\frac{I K}{A I}=\frac{r}{2 x}=\frac{1}{2}$.
Then $\angle B A I=30^{\circ}, \angle B A C=60^{\circ}$
$\Rightarrow \angle B H C=\angle B I C=\angle B O C=120^{\circ}$
$\Rightarrow B, H, I, C, B, H, O, C$ are concyclic
$\Rightarrow B, H, I, O, C$ are concyclic.
Hence $\angle H I O=180^{\circ}-\angle H B O$
$$
\begin{array}{l}
=180^{\circ}-\left(\left(90^{\circ}-\angle C\right)-\left(90^{\circ}-\angle A\right)\right) \\
=180^{\circ}-\left(60^{\circ}-\angle C\right)=120^{\circ}+\angle C \\
=120^{\circ}+\left(120^{\circ}-\angle B\right) \\
=240^{\circ}-\angle B .
\end{array}
$$