Given a regular triangle $ OAB$ with side length $ 1$. Let $ M,\ N$ be the point which divides equally $ AB$ into three parts, $ M$ is the closest to $ A$.
Denote $ \overrightarrow{OA} \equal{} \overrightarrow{a},\ \overrightarrow{OM} \equal{} \overrightarrow{m},\ \overrightarrow{ON} \equal{} \overrightarrow{n},\ \overrightarrow{OB} \equal{} \overrightarrow{b}$.

(1) Find the real numbers $ x,\ y$ such that $ \overrightarrow{m} \equal{} x \overrightarrow{a} \plus{} y \overrightarrow{b}$.

(2) Find the dot product of $ \overrightarrow{m},\ \overrightarrow{n}$ denoted by $ (\overrightarrow{m},\ \overrightarrow{n})$.
Given a regular triangle \( OAB \) with side length \( 1 \). Let \( M \) and \( N \) be the points that divide \( AB \) into three equal parts, with \( M \) being the closest to \( A \). We need to find the real numbers \( x \) and \( y \) such that \( \overrightarrow{m} = x \overrightarrow{a} + y \overrightarrow{b} \), and then find the dot product of \( \overrightarrow{m} \) and \( \overrightarrow{n} \).

1. **Finding \( \overrightarrow{m} \) in terms of \( \overrightarrow{a} \) and \( \overrightarrow{b} \):**

   Since \( M \) divides \( AB \) into three equal parts, we can use the section formula. The section formula states that if a point \( P \) divides the line segment joining points \( A \) and \( B \) in the ratio \( m:n \), then the position vector of \( P \) is given by:
   \[
   \overrightarrow{P} = \frac{n \overrightarrow{A} + m \overrightarrow{B}}{m+n}
   \]

   Here, \( M \) divides \( AB \) in the ratio \( 1:2 \). Therefore,
   \[
   \overrightarrow{M} = \frac{2 \overrightarrow{A} + 1 \overrightarrow{B}}{1+2} = \frac{2 \overrightarrow{A} + \overrightarrow{B}}{3}
   \]

   Thus, \( x = \frac{2}{3} \) and \( y = \frac{1}{3} \).

2. **Finding \( \overrightarrow{n} \) in terms of \( \overrightarrow{a} \) and \( \overrightarrow{b} \):**

   Similarly, \( N \) divides \( AB \) in the ratio \( 2:1 \). Therefore,
   \[
   \overrightarrow{N} = \frac{1 \overrightarrow{A} + 2 \overrightarrow{B}}{1+2} = \frac{\overrightarrow{A} + 2 \overrightarrow{B}}{3}
   \]

3. **Finding the dot product \( (\overrightarrow{m}, \overrightarrow{n}) \):**

   We have:
   \[
   \overrightarrow{m} = \frac{2 \overrightarrow{A} + \overrightarrow{B}}{3}, \quad \overrightarrow{n} = \frac{\overrightarrow{A} + 2 \overrightarrow{B}}{3}
   \]

   The dot product is given by:
   \[
   (\overrightarrow{m}, \overrightarrow{n}) = \left( \frac{2 \overrightarrow{A} + \overrightarrow{B}}{3} \right) \cdot \left( \frac{\overrightarrow{A} + 2 \overrightarrow{B}}{3} \right)
   \]

   Simplifying, we get:
   \[
   (\overrightarrow{m}, \overrightarrow{n}) = \frac{1}{9} \left( (2 \overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} + 2 \overrightarrow{B}) \right)
   \]

   Expanding the dot product:
   \[
   (2 \overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} + 2 \overrightarrow{B}) = 2 \overrightarrow{A} \cdot \overrightarrow{A} + 4 \overrightarrow{A} \cdot \overrightarrow{B} + \overrightarrow{B} \cdot \overrightarrow{A} + 2 \overrightarrow{B} \cdot \overrightarrow{B}
   \]

   Since \( \overrightarrow{A} \cdot \overrightarrow{A} = 1 \) (magnitude of \( \overrightarrow{A} \) is 1), \( \overrightarrow{B} \cdot \overrightarrow{B} = 1 \) (magnitude of \( \overrightarrow{B} \) is 1), and \( \overrightarrow{A} \cdot \overrightarrow{B} = \overrightarrow{B} \cdot \overrightarrow{A} = \cos(120^\circ) = -\frac{1}{2} \) (since \( OAB \) is an equilateral triangle with \( 120^\circ \) between \( \overrightarrow{A} \) and \( \overrightarrow{B} \)):

   \[
   (2 \overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} + 2 \overrightarrow{B}) = 2 \cdot 1 + 4 \cdot \left( -\frac{1}{2} \right) + \left( -\frac{1}{2} \right) + 2 \cdot 1
   \]
   \[
   = 2 - 2 - \frac{1}{2} + 2 = 1.5
   \]

   Therefore,
   \[
   (\overrightarrow{m}, \overrightarrow{n}) = \frac{1.5}{9} = \frac{1}{6}
   \]

The final answer is \( \boxed{ \frac{1}{6} } \)