(1). $\frac{1}{6}$
(2). $\frac{13}{2}$

#### 解析：

$\because \overrightarrow{A D}=\lambda \overrightarrow{B C}, \therefore A D / / B C, \therefore \angle B A D=180^{\circ}-\angle B=120^{\circ}$,
\begin{aligned} &\overrightarrow{A B} \cdot \overrightarrow{A D}=\lambda \overrightarrow{B C} \cdot \overrightarrow{A B}=\lambda|\overrightarrow{B C}| \cdot|\overrightarrow{A B}| \\ &=\lambda \times 6 \times 3 \times\left(-\frac{1}{2}\right)=-9 \lambda=-\frac{3}{2} \end{aligned}

$$\because B C=6, \therefore C(6,0),$$
$\because|A B|=3, \angle A B C=60^{\circ}, \therefore A$ 的坐标为 $A\left(\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)$,
$$\because \text { 又 } \because \overrightarrow{A D}=\frac{1}{6} \overrightarrow{B C} \text {, 则 } D\left(\frac{5}{2}, \frac{3 \sqrt{3}}{2}\right) \text {, 设 } M(x, 0) \text {, 则 } N(x+1,0) \text { (其中 } 0 \leq x \leq 5 \text { ), }$$
$$\overrightarrow{D M}=\left(x-\frac{5}{2},-\frac{3 \sqrt{3}}{2}\right), \quad \overrightarrow{D N}=\left(x-\frac{3}{2},-\frac{3 \sqrt{3}}{2}\right) \text {, }$$
$$\overrightarrow{D M} \cdot \overrightarrow{D N}=\left(x-\frac{5}{2}\right)\left(x-\frac{3}{2}\right)+\left(\frac{3 \sqrt{3}}{2}\right)^{2}=x^{2}-4 x+\frac{21}{2}=(x-2)^{2}+\frac{13}{2},$$