II. (25 points) As shown in Figure 5, in $\triangle A B C$, $\angle C=$ $90^{\circ}$, $D$ is a point on side $B C$, and $\angle A D C=45^{\circ}$. Draw $D E \perp A B$ at point $E$, and $A E: B E=10: 3$. If $D E=\sqrt{2}$, try to find the length of the angle bisector $C F$ of $\angle C$.
Let $B E=3 x, A E=10 x$. By the Pythagorean theorem, it is easy to get
$$
\begin{array}{l}
B D=\sqrt{2+9 x^{2}}, A D=\sqrt{2+100 x^{2}}, \\
C D=A C=\frac{A D}{\sqrt{2}}=\sqrt{1+50 x^{2}} .
\end{array}
$$

From $S_{\triangle A B D}=\frac{1}{2} A B \cdot D E=\frac{1}{2} B D \cdot A C$, we get
$$
\frac{1}{2} \times 13 x \times \sqrt{2}=\frac{1}{2} \sqrt{2+9 x^{2}} \times \sqrt{1+50 x^{2}} \text {. }
$$

Squaring both sides and simplifying, we get
$$
450 x^{4}-229 x^{2}+2=0 \text {. }
$$

Solving, we get $x^{2}=\frac{229 \pm \sqrt{48841}}{900}=\frac{229 \pm 221}{900}=\frac{1}{2}$ or $\frac{2}{225}$. Thus, $x=\frac{\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{15}$.
Since $A B>B C$, we have
$$
13 x>\sqrt{2+9 x^{2}}+\sqrt{1+50 x^{2}}>\sqrt{2}+1 .
$$

Therefore, $x>\frac{\sqrt{2}+1}{13}>\frac{\sqrt{2}}{15}$. Hence, $x=\frac{\sqrt{2}}{2}$.
From $S_{\triangle B C F}+S_{\triangle A C F}=S_{\triangle A B C}$, we get
$$
\frac{1}{2}(B C+A C) C F \sin 45^{\circ}=\frac{1}{2} A C \cdot B C \text {, }
$$

which simplifies to $C F \cdot \frac{\sqrt{2}}{2}\left(\frac{3}{2} \sqrt{26}+\sqrt{26}\right)=\sqrt{26} \times \frac{3}{2} \times \sqrt{26}$. Solving, we get $C F=\frac{6}{5} \sqrt{13}$.