II. (25 points) As shown in Figure 3, $P Q$ is tangent to $\odot O$ at point $Q$, and $Q$ is the midpoint of $\widehat{A B}$. Chord $A C = B C$, the extension of $C A$ intersects $P Q$ at point $D$, $O E \perp B C$ at point $E$, connect $D E$, and $F$ is the midpoint of $D E$. Given that $Q F = 2$, $A C = 6$. Find the length of $O E$.
II. As shown in Figure 11, extend $PQ$ and $CB$ to intersect at $G$.

It is easy to see that $\overparen{QAC}$ is a semicircle. Connect $QC$, then $QC$ is the diameter of $\odot O$.
$$
\begin{array}{c}
\text { By } P Q \perp C Q, \\
\angle A C Q = \angle B C Q \\
\Rightarrow \triangle D C Q \cong \triangle G C Q \Rightarrow D Q = Q G .
\end{array}
$$
$$
\text { Also, } D F = F E \text {, so } E G = 2 F Q = 2 \times 2 = 4 \text {. }
$$

From $O E \perp C B$ we get
$$
E B = E C = \frac{1}{2} B C = \frac{1}{2} A C = \frac{1}{2} \times 6 = 3 \text {. }
$$

Thus, $G B = E G - E B = 4 - 3 = 1$.
By the secant-tangent theorem, we have
$$
G Q^{2} = G B \cdot G C = 1 \times (1 + 6) = 7 \Rightarrow G Q = \sqrt{7} \text {. }
$$

By the Pythagorean theorem, we get
$$
Q C = \sqrt{G C^{2} - G Q^{2}} = \sqrt{7^{2} - \sqrt{7}^{2}} = \sqrt{42} \text {. }
$$

It is easy to see that $\triangle C O E \sim \triangle C G Q$.
Therefore, $\frac{O E}{G Q} = \frac{C E}{C Q}$.
Thus, $\frac{O E}{\sqrt{7}} = \frac{3}{\sqrt{42}}$, which means $O E = \frac{\sqrt{6}}{2}$.