如图, 在 $\triangle A B C$ 中, $A B=2, A C=3, \angle B A C=60^{\circ}, \overrightarrow{D B}=2 \overrightarrow{A D}$,

$\overrightarrow{C E}=2 \overrightarrow{E B}$<ImageHere>

(1) 求 $C D$ 的长;

(2) 求 $\overrightarrow{A B} \cdot \overrightarrow{D E}$ 的值.
(1) $\because \overrightarrow{D B}=2 \overrightarrow{A D}, \therefore \overrightarrow{A D}=\frac{1}{3} \overrightarrow{A B}, \therefore \overrightarrow{C D}=\overrightarrow{A D}-\overrightarrow{A C}=\frac{1}{3} \overrightarrow{A B}-\overrightarrow{A C}$,

$\because A B=2, A C=3, \angle B A C=60^{\circ}, \therefore \overrightarrow{A B} \cdot \overrightarrow{A C}=|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \cos 60^{\circ}=2 \times 3 \times \frac{1}{2}=3$,

$\because \overrightarrow{C D}^{2}=\left(\frac{1}{3} \overrightarrow{A B}-\overrightarrow{A C}\right)^{2}=\frac{1}{9} \overrightarrow{A B}^{2}-\frac{2}{3} \overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{A C}^{2}=\frac{1}{9} \times 2^{2}-\frac{2}{3} \times 3+3^{2}$

$=\frac{67}{9}$,

$\therefore|\overrightarrow{C D}|=\sqrt{\frac{67}{9}}=\frac{\sqrt{67}}{3}$.

(2) $\because \overrightarrow{C E}=2 \overrightarrow{E B}, \quad \therefore \overrightarrow{B E}=\frac{1}{3} \overrightarrow{B C}$,

$\overrightarrow{D E}=\overrightarrow{D B}+\overrightarrow{B E}=\frac{2}{3} \overrightarrow{A B}+\frac{1}{3} \overrightarrow{B C}=\frac{2}{3} \overrightarrow{A B}+\frac{1}{3}(\overrightarrow{A C}-\overrightarrow{A B})=\frac{1}{3} \overrightarrow{A B}+\frac{1}{3} \overrightarrow{A C}$,

$\therefore \overrightarrow{A B} \cdot \overrightarrow{D E}=\overrightarrow{A B} \cdot\left(\frac{1}{3} \overrightarrow{A B}+\frac{1}{3} \overrightarrow{A C}\right)=\frac{1}{3} \overrightarrow{A B}^{2}+\frac{1}{3} \overrightarrow{A B} \cdot \overrightarrow{A C}=\frac{1}{3} \times 2^{2}+\frac{1}{3} \times 3=\frac{7}{3}$.