II. (50 points) As shown in Figure 3, given that $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$ respectively, $O$ and $I$ do not coincide, $I$ is inside or on the boundary of $\triangle AOC$, and either $O$ is inside $\triangle ABC$ or $AB=AC$.
Try to find a necessary and sufficient condition (expressed in terms of the interior angles of $\triangle ABC$) for the equation $\angle BAC - \angle BIO = 30^{\circ}$ to hold.
The required necessary and sufficient condition is
$\angle B A C=60^{\circ}$ or $100^{\circ}$.
Sufficiency.
As shown in Figure 7, if $\angle B A C=60^{\circ}$, then
$\angle B O C=2 \angle B A C=120^{\circ}$,
$\angle B I C=180^{\circ}-\frac{\angle A B C+\angle A C B}{2}$
$=90^{\circ}+\frac{\angle B A C}{2}=120^{\circ}$.
Thus, $\angle B O C=\angle B I C$.
Therefore, $B, O, I, C$ are concyclic.
Hence $\angle B I O=\angle O C B=\angle O B C$
$=90^{\circ}-\angle B A C=30^{\circ}$.
Thus, $\angle B A C-\angle B I O=30^{\circ}$.
If $\angle B A C=100^{\circ}$, then $O$ is outside $\triangle A B C$, as shown in Figure 8. By the problem, it must be that $A B=A C$. By symmetry, $A, O, I$ are collinear, $A O \perp B C$, and $I$ lies on segment $A O$. Clearly,
$$
\begin{array}{l}
\angle B I O=\frac{\angle B A C}{2}+\frac{\angle A B C}{2} \\
=\frac{\angle B A C}{2}+\frac{1}{2} \cdot \frac{180^{\circ}-\angle B A C}{2}=70^{\circ} .
\end{array}
$$

Thus, $\angle B A C-\angle B I O=30^{\circ}$.
Necessity.
Assume $\angle B A C-\angle B I O=30^{\circ}, I$ is inside or on the boundary of $\triangle A O C$. We discuss two cases below.
(1) If $A B=A C$, then by symmetry, $A, O, I$ are collinear, $A O \perp B C$. Since $I$ is inside or on the boundary of $\triangle A O C$, $I$ must lie on segment $A O$ (as shown in Figure 8).
Thus, $\angle B I O=\frac{\angle B A C}{2}+\frac{\angle A B C}{2}$.
Substituting $\angle B A C-\angle B I O=30^{\circ}$, we solve to get
$\angle B A C=100^{\circ}$.
(2) If $O$ is inside $\triangle A B C$, then $\triangle A B C$ is an acute triangle. We prove: $\angle B I O=30^{\circ}$, thus, $\angle B A C=30^{\circ}+\angle B I O=60^{\circ}$. For this, we discuss three cases.
(i) If $\angle B I O \geqslant 60^{\circ}$, then
$\angle B A C=30^{\circ}+\angle B I O \geqslant 90^{\circ}$, which contradicts $\triangle A B C$ being an acute triangle.
(ii) If $\angle B I O \in\left(30^{\circ}, 60^{\circ}\right)$, then
$$
\begin{array}{l}
\angle B A C=30^{\circ}+\angle B I O \in\left(60^{\circ}, 90^{\circ}\right), \\
\angle O C B=90^{\circ}-\angle B A C \in\left(0^{\circ}, 30^{\circ}\right) .
\end{array}
$$

Thus, $\angle B I O>\angle O C B$.
Therefore, $I$ is inside the circumcircle of $\triangle B O C$ (as shown in Figure 7). Hence $\angle B I C>\angle B O C$. Solving, we get $\angle B A C<60^{\circ}$, which contradicts $\angle B A C \in\left(60^{\circ}, 90^{\circ}\right)$.
(iii) If $\angle B I O \in\left(0^{\circ}, 30^{\circ}\right)$, similarly to (ii), we derive $\angle B A C \in\left(30^{\circ}, 60^{\circ}\right)$, which is a contradiction.
Thus, $\angle B I O=30^{\circ}, \angle B A C=30^{\circ}+\angle B I O=60^{\circ}$.
In summary, a necessary and sufficient condition for $\angle B A C-\angle B I O=30^{\circ}$ is $\angle B A C=60^{\circ}$ or $100^{\circ}$.