II. (25 points) As shown in Figure 4, given that $AB$ is the diameter of $\odot O$, chord $CD$ intersects $AB$ at point $E$. A tangent to $\odot O$ is drawn from point $A$ and intersects the extension of $CD$ at point $F$. Given $AC=8$, $CE:ED=6:5$, and $AE:EB=2:3$. Find the length of $AB$ and the value of $\tan \angle ECB$.

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Let $C E=6 x, E D=5 x, A E=2 y$, $E B=3 y, D F=z$.
By the intersecting chords theorem, we have
$$
A E \cdot B E=C E \cdot E D \Rightarrow y=\sqrt{5} x \text {. }
$$

By the secant-tangent theorem, we have
$$
A F^{2}=D F \cdot C F=z(z+11 x) \text {. }
$$

By the Pythagorean theorem, we have
$$
\begin{array}{l}
A F^{2}=E F^{2}-A E^{2} \\
\Rightarrow z(z+11 x)=(5 x+z)^{2}-(2 y)^{2} .
\end{array}
$$

Substituting $y=\sqrt{5} x$ yields $z=5 x$, meaning $D$ is the midpoint of $E F$.
Therefore, $A D=D E=5 x$.
It is easy to see that $B C=B E=3 \sqrt{5} x, A B=5 \sqrt{5} x$.
$$
\begin{array}{l}
\text { Also, } A B^{2}-B C^{2}=A C^{2} \\
\Rightarrow(5 \sqrt{5} x)^{2}-(3 \sqrt{5} x)^{2}=8^{2} \\
\Rightarrow x=\frac{2 \sqrt{5}}{5} \Rightarrow A B=10 .
\end{array}
$$

Since $\angle E C B=\angle E A D=\angle D E A$, we have
$$
\tan \angle E C B=\tan \angle D E A=\frac{A F}{A E}=2 .
$$