Let $m,\ n$ be the given distinct positive integers. Answer the following questions.

(1) Find the real number $\alpha \ (|\alpha |<1)$ such that $\int_{-\pi}^{\pi}\sin (m+\alpha )x\ \sin (n+\alpha )x\ dx=0$.

(2) Find the real number $\beta$ satifying the sytem of equation $\int_{-\pi}^{\pi}\sin^{2}(m+\beta )x\ dx=\pi+\frac{2}{4m-1}$, $\int_{-\pi}^{\pi}\sin^{2}(n+\beta )x\ dx=\pi+\frac{2}{4n-1}$.
### Part 1:
We need to find the real number $\alpha$ such that:
\[
\int_{-\pi}^{\pi} \sin((m+\alpha)x) \sin((n+\alpha)x) \, dx = 0
\]

1. **Use the product-to-sum identities:**
   \[
   \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]
   \]
   Applying this identity, we get:
   \[
   \sin((m+\alpha)x) \sin((n+\alpha)x) = \frac{1}{2} [\cos((m+\alpha)x - (n+\alpha)x) - \cos((m+\alpha)x + (n+\alpha)x)]
   \]
   Simplifying the arguments of the cosines:
   \[
   \sin((m+\alpha)x) \sin((n+\alpha)x) = \frac{1}{2} [\cos((m-n)x) - \cos((m+n+2\alpha)x)]
   \]

2. **Integrate each term separately:**
   \[
   \int_{-\pi}^{\pi} \sin((m+\alpha)x) \sin((n+\alpha)x) \, dx = \frac{1}{2} \left( \int_{-\pi}^{\pi} \cos((m-n)x) \, dx - \int_{-\pi}^{\pi} \cos((m+n+2\alpha)x) \, dx \right)
   \]

3. **Evaluate the integrals:**
   - For $\int_{-\pi}^{\pi} \cos((m-n)x) \, dx$:
     \[
     \int_{-\pi}^{\pi} \cos((m-n)x) \, dx = \left. \frac{\sin((m-n)x)}{m-n} \right|_{-\pi}^{\pi} = \frac{\sin((m-n)\pi) - \sin(-(m-n)\pi)}{m-n} = 0
     \]
     because $\sin(k\pi) = 0$ for any integer $k$.

   - For $\int_{-\pi}^{\pi} \cos((m+n+2\alpha)x) \, dx$:
     \[
     \int_{-\pi}^{\pi} \cos((m+n+2\alpha)x) \, dx = \left. \frac{\sin((m+n+2\alpha)x)}{m+n+2\alpha} \right|_{-\pi}^{\pi} = \frac{\sin((m+n+2\alpha)\pi) - \sin(-(m+n+2\alpha)\pi)}{m+n+2\alpha} = 0
     \]
     because $\sin(k\pi) = 0$ for any integer $k$.

4. **Conclusion:**
   Since both integrals are zero, the original integral is zero for any $\alpha$ such that $|\alpha| < 1$.

### Part 2:
We need to find the real number $\beta$ satisfying the system of equations:
\[
\int_{-\pi}^{\pi} \sin^2((m+\beta)x) \, dx = \pi + \frac{2}{4m-1}
\]
\[
\int_{-\pi}^{\pi} \sin^2((n+\beta)x) \, dx = \pi + \frac{2}{4n-1}
\]

1. **Use the identity for $\sin^2 x$:**
   \[
   \sin^2 x = \frac{1 - \cos(2x)}{2}
   \]
   Applying this identity, we get:
   \[
   \sin^2((m+\beta)x) = \frac{1 - \cos(2(m+\beta)x)}{2}
   \]

2. **Integrate each term separately:**
   \[
   \int_{-\pi}^{\pi} \sin^2((m+\beta)x) \, dx = \int_{-\pi}^{\pi} \frac{1 - \cos(2(m+\beta)x)}{2} \, dx
   \]
   \[
   = \frac{1}{2} \int_{-\pi}^{\pi} 1 \, dx - \frac{1}{2} \int_{-\pi}^{\pi} \cos(2(m+\beta)x) \, dx
   \]

3. **Evaluate the integrals:**
   - For $\int_{-\pi}^{\pi} 1 \, dx$:
     \[
     \int_{-\pi}^{\pi} 1 \, dx = 2\pi
     \]

   - For $\int_{-\pi}^{\pi} \cos(2(m+\beta)x) \, dx$:
     \[
     \int_{-\pi}^{\pi} \cos(2(m+\beta)x) \, dx = \left. \frac{\sin(2(m+\beta)x)}{2(m+\beta)} \right|_{-\pi}^{\pi} = \frac{\sin(2(m+\beta)\pi) - \sin(-2(m+\beta)\pi)}{2(m+\beta)} = 0
     \]
     because $\sin(k\pi) = 0$ for any integer $k$.

4. **Combine the results:**
   \[
   \int_{-\pi}^{\pi} \sin^2((m+\beta)x) \, dx = \frac{1}{2} (2\pi) - \frac{1}{2} (0) = \pi
   \]

5. **Solve the system of equations:**
   \[
   \pi = \pi + \frac{2}{4m-1}
   \]
   \[
   \pi = \pi + \frac{2}{4n-1}
   \]
   This implies:
   \[
   \frac{2}{4m-1} = 0 \quad \text{and} \quad \frac{2}{4n-1} = 0
   \]
   which is a contradiction since $m$ and $n$ are positive integers. Therefore, there is no real number $\beta$ that satisfies both equations.

The final answer is $\boxed{\alpha \text{ such that } |\alpha| < 1}$ for part 1 and $\boxed{\text{No solution}}$ for part 2.