Find all positive integers $a,b$ with $a>1$ such that,  $b$ is a divisor of $a-1$ and $2a+1$ is a divisor of $5b-3$.
To find all positive integers \(a, b\) with \(a > 1\) such that \(b\) is a divisor of \(a-1\) and \(2a+1\) is a divisor of \(5b-3\), we proceed as follows:

1. **Given Conditions:**
   - \(b \mid (a-1)\)
   - \(2a+1 \mid (5b-3)\)

2. **Inequality Analysis:**
   - Since \(b \mid (a-1)\), we can write \(a-1 = kb\) for some integer \(k\).
   - Therefore, \(a = kb + 1\).

3. **Divisibility Condition:**
   - We need \(2a+1 \mid (5b-3)\).
   - Substituting \(a = kb + 1\) into \(2a+1\), we get:
     \[
     2a + 1 = 2(kb + 1) + 1 = 2kb + 2 + 1 = 2kb + 3
     \]
   - Thus, \(2kb + 3 \mid 5b - 3\).

4. **Bounding \(5b-3\):**
   - We know \(5b - 3 < 5a - 3\).
   - Since \(a = kb + 1\), we have:
     \[
     5a - 3 = 5(kb + 1) - 3 = 5kb + 5 - 3 = 5kb + 2
     \]
   - Therefore, \(5b - 3 < 5kb + 2\).

5. **Possible Values for \(5b-3\):**
   - Since \(2kb + 3 \mid 5b - 3\), we consider the possible values:
     \[
     5b - 3 = m(2kb + 3) \quad \text{for some integer } m
     \]
   - We need to check the cases \(m = 1\) and \(m = 2\) since higher values would make \(5b - 3\) too large.

6. **Case 1: \(m = 1\):**
   - \(5b - 3 = 2kb + 3\)
   - Solving for \(b\):
     \[
     5b - 2kb = 6 \implies b(5 - 2k) = 6
     \]
   - Possible integer solutions for \(b\) and \(k\) are:
     - \(b = 2, k = 2\) (since \(5 - 2k = 1\))
     - \(b = 6, k = 1\) (since \(5 - 2k = 3\))

7. **Case 2: \(m = 2\):**
   - \(5b - 3 = 2(2kb + 3)\)
   - Solving for \(b\):
     \[
     5b - 3 = 4kb + 6 \implies 5b - 4kb = 9 \implies b(5 - 4k) = 9
     \]
   - Possible integer solutions for \(b\) and \(k\) are:
     - \(b = 9, k = 1\) (since \(5 - 4k = 1\))

8. **Verification:**
   - For \(b = 2, k = 2\):
     \[
     a = kb + 1 = 2 \cdot 2 + 1 = 5
     \]
     - Check \(2a + 1 = 11\) and \(5b - 3 = 7\), not a divisor.
   - For \(b = 6, k = 1\):
     \[
     a = kb + 1 = 1 \cdot 6 + 1 = 7
     \]
     - Check \(2a + 1 = 15\) and \(5b - 3 = 27\), not a divisor.
   - For \(b = 9, k = 1\):
     \[
     a = kb + 1 = 1 \cdot 9 + 1 = 10
     \]
     - Check \(2a + 1 = 21\) and \(5b - 3 = 42\), \(21 \mid 42\).

Thus, the only valid solution is \(a = 10\) and \(b = 9\).

The final answer is \(\boxed{(a, b) = (10, 9)}\).