Three. (25 points) As shown in Figure 4, given the line $l: y = kx + 2 (k < 0)$ intersects the $y$-axis at point $A$ and the $x$-axis at point $B$. The circle $\odot P$ with $OA$ as its diameter intersects $l$ at another point $D$. When the minor arc $\overparen{A D}$ is flipped along the line $l$, it intersects $OA$ at point $E$.
(1) When $k = -2$, find the length of $OE$;
(2) Does there exist a real number $k (k < 0)$ such that the arc obtained by flipping $\overparen{A D}$ along the line $l$ is tangent to $OA$? If it exists, find the value of $k$; if not, explain the reason.
Three, (1) As shown in Figure 8, connect
$O D, D E$. Then
$$
\begin{array}{l}
\angle D E O \\
=\angle E A D+\angle A D E \\
=\frac{\overparen{A E D^{\circ}}}{2}=\frac{\overparen{A m D}^{\circ}}{2} \\
=\angle A O D .
\end{array}
$$

Therefore, $O D=D E$.
When $k=-2$, it is easy to get $A(0,2), B(1,0), O A=2, O B$ $=1$. Then $A B=\sqrt{5}$.
Since $B O$ is tangent to $\odot P$ at point $O$, by the secant-tangent theorem, we have
$$
O B^{2}=B D \cdot A B \Rightarrow B D=\frac{\sqrt{5}}{5} \text {. }
$$

Draw $D C \perp A O$ at point $C$, then
$$
O E=2 O C, D C / / O B \text {. }
$$

Thus, $\frac{O C}{O A}=\frac{B D}{B A} \Rightarrow \frac{O C}{2}=\frac{\frac{\sqrt{5}}{5}}{\sqrt{5}} \Rightarrow O C=\frac{2}{5}$. Hence $O E=\frac{4}{5}$.
(2) Assume there exists a real number $k(k<0)$ such that the arc $\overparen{A D}$, when flipped over line $l$, results in an arc that is tangent to $O A$, then the point of tangency must be $A$, i.e., $E$ coincides with $A$. From (1), we know $O D=A D$.
Also, $\angle A D O=90^{\circ}$, so,
$\angle O A D=45^{\circ}$.
At this point, $O B=O A=2, B(2,0)$.
Substituting $x=2, y=0$ into the equation $y=k x+2$, we get
$$
k=-1 \text {. }
$$

Therefore, there exists $k=-1$ such that the arc $\overparen{A D}$, when flipped over line $l$, results in an arc that is tangent to $O A$.
(Deng Jichun, No. 10 Middle School of Wugang City, Hunan Province, 422400)