$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dots+\frac{1}{2014\cdot2015}=\frac{m}{n},$$ where $\frac{m}{n}$ is irreducible. 
a) Find $m+n.$
b) Find the remainder of division of $(m+3)^{1444}$ to $n{}$.
1. **Simplify the given series using partial fractions:**
   \[
   \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}
   \]
   This is a telescoping series. Applying this to the given series:
   \[
   \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{2014 \cdot 2015}
   \]
   becomes:
   \[
   \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{2014} - \frac{1}{2015} \right)
   \]

2. **Observe the telescoping nature of the series:**
   Most terms cancel out:
   \[
   \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{2014} - \frac{1}{2015} \right)
   \]
   The series simplifies to:
   \[
   1 - \frac{1}{2015}
   \]

3. **Combine the remaining terms:**
   \[
   1 - \frac{1}{2015} = \frac{2015}{2015} - \frac{1}{2015} = \frac{2015 - 1}{2015} = \frac{2014}{2015}
   \]
   Thus, \( \frac{m}{n} = \frac{2014}{2015} \), where \( m = 2014 \) and \( n = 2015 \).

4. **Calculate \( m + n \):**
   \[
   m + n = 2014 + 2015 = 4029
   \]

5. **Find the remainder of \( (m+3)^{1444} \) modulo \( n \):**
   We need to find \( 2017^{1444} \mod 2015 \).

6. **Use Euler's theorem:**
   Euler's theorem states that if \( a \) and \( n \) are coprime, then:
   \[
   a^{\phi(n)} \equiv 1 \pmod{n}
   \]
   where \( \phi \) is the Euler's totient function. First, calculate \( \phi(2015) \):
   \[
   2015 = 5 \times 13 \times 31
   \]
   \[
   \phi(2015) = \phi(5) \times \phi(13) \times \phi(31) = (5-1)(13-1)(31-1) = 4 \times 12 \times 30 = 1440
   \]

7. **Simplify the exponent using Euler's theorem:**
   Since \( 2017 \equiv 2 \pmod{2015} \):
   \[
   2017^{1444} \equiv 2^{1444} \pmod{2015}
   \]
   Using \( \phi(2015) = 1440 \):
   \[
   2^{1444} = 2^{1440 + 4} = (2^{1440})(2^4) \equiv 1 \cdot 2^4 \pmod{2015} = 16
   \]

The final answer is \( \boxed{16} \)