find all positive integer $a\geq 2 $ and $b\geq2$ such that $a$ is even and all the digits of $a^b+1$ are equals.
We need to find all positive integers \(a \geq 2\) and \(b \geq 2\) such that \(a\) is even and all the digits of \(a^b + 1\) are equal.

1. **Representation of \(a^b + 1\)**:
   Let \(a^b + 1 = \overline{\underbrace{mm \dots m}_{k \text{ digits of } m}}\), where \(m\) and \(k\) are positive integers and \(m \leq 9\).

2. **Case 1: \(k = 1\)**:
   - If \(k = 1\), then \(a^b + 1 = m\), where \(m\) is a single digit.
   - We need to check for \(a\) and \(b\) such that \(a^b + 1\) is a single digit.
   - By checking small values, we find:
     - \(2^2 + 1 = 5\)
     - \(2^3 + 1 = 9\)
   - Therefore, \((a, b) = (2, 2)\) and \((a, b) = (2, 3)\) are solutions for this case.

3. **Case 2: \(k \geq 2\)**:
   - We need to consider the number \(a^b + 1\) modulo 4.
   - Since \(a\) is even, \(a^b\) is divisible by 4, so \(a^b \equiv 0 \pmod{4}\).
   - Thus, \(a^b + 1 \equiv 1 \pmod{4}\).
   - The number \(\overline{\underbrace{mm \dots m}_{k \text{ digits of } m}}\) must also be congruent to 1 modulo 4.
   - This implies \(\overline{mm} \equiv 1 \pmod{4}\).
   - Since \(\overline{mm} = 11m\), we have \(11m \equiv 1 \pmod{4}\).
   - Solving \(11m \equiv 1 \pmod{4}\), we get \(m \equiv 3 \pmod{4}\), so \(m = 3\) or \(m = 7\).

4. **Subcase 1: \(m = 3\)**:
   - Assume \(k \geq 3\).
   - Consider modulo 8:
     - \(\overline{\underbrace{33 \dots 3}_{k \text{ digits of } 3}} \equiv 333 \equiv 5 \pmod{8}\).
     - Thus, \(a^b + 1 \equiv 5 \pmod{8}\), so \(a^b \equiv 4 \pmod{8}\).
     - If \(b \geq 3\), then \(8 \mid a^b\), which is not possible since \(a^b \equiv 4 \pmod{8}\).
     - Therefore, \(b = 2\).
   - Consider modulo 3:
     - \(a^2 + 1 \equiv \overline{\underbrace{33 \dots 3}_{k \text{ digits of } 3}} \equiv 0 \pmod{3}\).
     - Thus, \(a^2 \equiv -1 \pmod{3}\), which is impossible.
   - Therefore, \(k = 2\).
   - Checking small values, we find:
     - \(2^5 + 1 = 33\).
   - Therefore, \((a, b) = (2, 5)\) is a solution for this subcase.

5. **Subcase 2: \(m = 7\)**:
   - Assume \(k \geq 5\).
   - Consider modulo 32:
     - \(\overline{\underbrace{77 \dots 7}_{k \text{ digits of } 7}} \equiv 77777 \equiv 17 \pmod{32}\).
     - Thus, \(a^b + 1 \equiv 17 \pmod{32}\), so \(a^b \equiv 16 \pmod{32}\).
     - If \(b \geq 5\), then \(32 \mid a^b\), which is not possible since \(a^b \equiv 16 \pmod{32}\).
     - Therefore, \(b = 2, 3, 4\).
   - Consider modulo 7:
     - \(a^2 + 1 \equiv \overline{\underbrace{77 \dots 7}_{k \text{ digits of } 7}} \equiv 0 \pmod{7}\).
     - Thus, \(a^2 \equiv -1 \pmod{7}\), which is impossible.
     - Similarly, \(a^4 + 1 \equiv 0 \pmod{7}\), so \(a^4 \equiv -1 \pmod{7}\), which is also impossible.
   - Consider \(a^3\):
     - \(16 \mid a^3\) and \(32 \nmid a^3\), so \(a^3 = 2^4 \cdot n\), where \(n\) is an odd number, which is impossible.
   - Therefore, \(k \leq 4\).
   - Checking small values, we find:
     - \(6^5 + 1 = 77777\).
   - Therefore, \((a, b) = (6, 5)\) is a solution for this subcase.

6. **Conclusion**:
   - Combining all cases and subcases, the solutions are \((a, b) = (2, 2), (2, 3), (2, 5), (6, 5)\).

The final answer is \(\boxed{(2, 2), (2, 3), (2, 5), (6, 5)}\).