2. Given that $\triangle A B C$ is inscribed in $\odot O$ with radius 1, and $A B \cdot A C=A B-A C=1$. Try to find the measures of the three interior angles of $\triangle A B C$, and prove your conclusion.
2. From $A B+A C=\sqrt{(A B-A C)^{2}+4 A B \cdot A C}=\sqrt{5}$, we solve to get $A B=\frac{\sqrt{5}+1}{2}, A C=\frac{\sqrt{5}-1}{2}$.
Take a point $D$ on $O A$ such that $O D=A C=\frac{\sqrt{5}-1}{2}$, then $A D=1-O D=\frac{3-\sqrt{5}}{2}$.
In $\triangle O A C$ and $\triangle C A D$, we have
$$
\angle O A C=\angle C A D, \frac{A D}{A C}=\frac{\sqrt{5}-1}{2}=\frac{A C}{O A} \text {. }
$$

Therefore, $\triangle O A C \backsim \triangle C A D$.
Thus, $C D=A C=O D$,
$\angle A C D=\angle A O C=\angle O C D$,
$\angle O A C=\angle O C A=2 \angle A O C$.
Hence $5 \angle A O C=180^{\circ} \Rightarrow \angle A O C=36^{\circ}$
$\Rightarrow \angle A B C=18^{\circ}$.
Take a point $E$ on $A B$ such that $A E=1$, then
$$
E B=A B-1=\frac{\sqrt{5}-1}{2} \text {. }
$$

In $\triangle E B O$ and $\triangle O B A$, we have
$$
\angle E B O=\angle O B A, \frac{O B}{B E}=\frac{\sqrt{5}+1}{2}=\frac{A B}{O B} .
$$

Thus, $\triangle E B O \backsim \triangle O B A$.
Therefore, $B E=O E, O E=\frac{\sqrt{5}-1}{2}$.
Hence, $\angle O A E=36^{\circ}, \angle O B A=36^{\circ}$,
$\angle A O B=180^{\circ}-36^{\circ}-36^{\circ}=108^{\circ}$.
Therefore, $\angle A C B=54^{\circ}$ or $126^{\circ}, \angle B A C=108^{\circ}$ or $36^{\circ}$.
Thus, in $\triangle A B C$,
$\angle A B C=18^{\circ}, \angle A C B=54^{\circ}, \angle B A C=108^{\circ}$
or $\angle A B C=18^{\circ}, \angle A C B=126^{\circ}, \angle B A C=36^{\circ}$.