For any positive integer $m\ge2$ define  $A_m=\{m+1, 3m+2, 5m+3, 7m+4, \ldots, (2k-1)m + k, \ldots\}$.

(1) For every $m\ge2$, prove that there exists a positive integer $a$ that satisfies $1\le a<m$ and $2^a\in A_m$ or $2^a+1\in A_m$.

(2) For a certain $m\ge2$, let $a, b$ be positive integers that satisfy $2^a\in A_m$, $2^b+1\in A_m$. Let $a_0, b_0$ be the least such pair $a, b$. 
Find the relation between $a_0$ and $b_0$.
### Part (1)

To prove that for every \( m \ge 2 \), there exists a positive integer \( a \) that satisfies \( 1 \le a < m \) and \( 2^a \in A_m \) or \( 2^a + 1 \in A_m \):

1. **Expression for \( 2^a \in A_m \)**:
   - We need to find a positive integer \( k \) such that \( 2^a = (2k-1)m + k \).
   - This can be rewritten as:
     \[
     2^a = (2m+1)k - m
     \]
   - Rearranging, we get:
     \[
     2^a + m \equiv 0 \pmod{2m+1}
     \]
   - This implies:
     \[
     2^{a+1} + 2m \equiv 0 \pmod{2m+1}
     \]
   - Therefore:
     \[
     2^{a+1} \equiv 1 \pmod{2m+1}
     \]

2. **Existence of \( a \)**:
   - By Euler's theorem, \( 2^{\varphi(2m+1)} \equiv 1 \pmod{2m+1} \).
   - Thus, we can take \( a = \varphi(2m+1) - 1 \geq 1 \).
   - Let \( a_0 \) be the least such \( a \), then:
     \[
     2^{a_0+1} \equiv 1 \pmod{2m+1}
     \]

3. **Expression for \( 2^b + 1 \in A_m \)**:
   - We need to find a positive integer \( \ell \) such that \( 2^b + 1 = (2\ell-1)m + \ell \).
   - This can be rewritten as:
     \[
     2^b + 1 = (2m+1)\ell - m
     \]
   - Rearranging, we get:
     \[
     2^b + m + 1 \equiv 0 \pmod{2m+1}
     \]
   - This implies:
     \[
     2^{b+1} + 2m + 2 \equiv 0 \pmod{2m+1}
     \]
   - Therefore:
     \[
     2^{b+1} \equiv -1 \pmod{2m+1}
     \]

4. **Existence of \( b \)**:
   - Let \( b_0 \) be the least such \( b \), then:
     \[
     2^{b_0+1} \equiv -1 \pmod{2m+1}
     \]

### Part (2)

To find the relation between \( a_0 \) and \( b_0 \):

1. **Contradiction if \( a_0 < b_0 \)**:
   - If \( a_0 < b_0 \), then:
     \[
     2^{(b_0-a_0-1)+1} \equiv -1 \pmod{2m+1}
     \]
   - This contradicts the minimality of \( b_0 \).

2. **Contradiction if \( b_0 < a_0 < 2b_0+1 \)**:
   - If \( b_0 < a_0 < 2b_0+1 \), then:
     \[
     2^{(a_0-b_0-1)+1} \equiv -1 \pmod{2m+1}
     \]
   - This again contradicts the minimality of \( b_0 \).

3. **Conclusion**:
   - We clearly have:
     \[
     2^{2(b_0+1)} \equiv 1 \pmod{2m+1}
     \]
   - Thus:
     \[
     a_0 = 2b_0 + 1
     \]

### Justification for \( m \ge 2 \):

1. **Case when \( a_0 < \varphi(2m+1)-1 \)**:
   - If \( a_0 < \varphi(2m+1)-1 \), then:
     \[
     a_0 \leq \frac{1}{2}\varphi(2m+1) - 1 \leq m-1
     \]
   - This satisfies \( 1 \le a < m \).

2. **Case when \( a_0 = \varphi(2m+1)-1 \)**:
   - If \( 2m+1 \) has at least two distinct prime factors, then:
     \[
     2m+1 = pq
     \]
   - Using properties of Euler's totient function and multiplicative order, we get a contradiction unless \( 2m+1 \) is a power of a prime.

3. **Conclusion**:
   - For \( m \ge 2 \), the relation \( a_0 = 2b_0 + 1 \) holds, and the restriction \( m \ge 2 \) avoids exceptions.

\(\blacksquare\)

The final answer is \( \boxed{ a_0 = 2b_0 + 1 } \)