12. In $\triangle A B C$, $\angle B=30^{\circ}, \angle C=45^{\circ}$.
(1) If $A B=2$, find the length of $B C$;
(2) If point $D$ is on side $B C$, and point $P$ satisfies $P A \perp A D$, take point $E$ on segment $A P$, and take point $F$ on the extension of $D A$ such that $A E=A F, D E=P F$.
(i) As shown in Figure 7, when point $P$ exactly falls on side $A B$, find the value of $\frac{C D}{B P}$;
(ii) As shown in Figure 8, when point $P$ exactly falls on the perpendicular bisector of segment $A B$ (and not on side $A B$), please directly write the value of $\frac{A C}{B P}$.
12. (1) Draw $A H \perp B C$, with the foot of the perpendicular at $H$. Then, in the right triangle $\triangle A H B$, by the problem, we have $A H=1, B H=\sqrt{3}$.
In the right triangle $\triangle A H C$, by the problem, we have $H C=A H=1 \Rightarrow B C=B H+H C=\sqrt{3}+1$.
(2) (i) As shown in Figure 10, when point $P$ is on side $A B$, draw $A H \perp B C$, with the foot of the perpendicular at $H$.

Let $A H=h$. Then $C H=h, A B=2 h$.
By $A P \perp A D \Rightarrow \angle D A H=\angle B=30^{\circ}$. In the right triangle $\triangle A H D$, $D H=\frac{\sqrt{3}}{3} h, A D=\frac{2 \sqrt{3}}{3} h$. Thus, $C D=C H-D H=h-\frac{\sqrt{3}}{3} h=\left(1-\frac{\sqrt{3}}{3}\right) h$. Also, $A E=A F, D E=P F$, $\angle D A E=\angle P A F=90^{\circ}$, so $\triangle D A E \cong \triangle P A F \Rightarrow A P=A D=\frac{2 \sqrt{3}}{3} h$ $\Rightarrow B P=A B-A P=2\left(1-\frac{\sqrt{3}}{3}\right) h$.
Therefore, $\frac{C D}{B P}=\frac{1}{2}$.
(ii) $\frac{A C}{B P}=\frac{1+\sqrt{3}}{2}$.