Find all pairs of natural numbers $(a,b)$ satisfying the following conditions:
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[*]$b-1$ is divisible by $a+1$ and
[*]$a^2+a+2$ is divisible by $b$.
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1. **Substitute \( t = a + 1 \):**
   - Given \( b - 1 \) is divisible by \( a + 1 \), we set \( t = a + 1 \). Thus, \( t \mid b - 1 \).
   - Given \( a^2 + a + 2 \) is divisible by \( b \), we need \( b \mid t^2 - t + 2 \) because \( t = a + 1 \) implies \( a = t - 1 \).

2. **Express \( b \) in terms of \( t \):**
   - Since \( t \mid b - 1 \), we can write \( b = tk + 1 \) for some \( k \in \mathbb{N} \).

3. **Substitute \( b = tk + 1 \) into the divisibility condition:**
   - We need \( tk + 1 \mid t^2 - t + 2 \).
   - Since \( tk + 1 \mid t^2 - t + 2 \), it follows that \( tk + 1 \mid 2tk + 2 \) (multiplying by \( k \) and adding 1).

4. **Simplify the divisibility condition:**
   - We have \( tk + 1 \mid t^2 - t + 2 \) and \( tk + 1 \mid 2tk + 2 \).
   - Therefore, \( tk + 1 \mid t^2 - t + 2 - (2tk + 2) = t^2 - t - 2tk - 2 \).
   - This simplifies to \( tk + 1 \mid t(t - 2k - 1) \).

5. **Analyze the divisibility:**
   - Since \( tk + 1 \) and \( t \) are coprime, \( tk + 1 \mid t - 2k - 1 \).

6. **Claim \( t = 2k + 1 \):**
   - Suppose \( t > 2k + 1 \). Then \( t - 2k - 1 \ge tk + 1 \), which implies \( t \ge tk + 2k + 2 \), leading to a contradiction since \( t \le 1 \) is not possible as \( t = a + 1 \ge 2 \).
   - Suppose \( 2k + 1 > t \). Then \( 2k + 1 \ge tk + t + 1 \), which implies \( t \le 1 \), again a contradiction.

7. **Conclude \( t = 2k + 1 \):**
   - Therefore, \( t = 2k + 1 \).

8. **Find \( a \) and \( b \):**
   - Since \( t = a + 1 \), we have \( a = t - 1 = 2k + 1 - 1 = 2k \).
   - Since \( b = tk + 1 \), we have \( b = (2k + 1)k + 1 = 2k^2 + k + 1 \).

Thus, the pairs \((a, b)\) are given by \((2k, 2k^2 + 2k + 1)\) for \( k \in \mathbb{N} \).

The final answer is \( \boxed{ (2k, 2k^2 + 2k + 1) } \) for \( k \in \mathbb{N} \).