Find all integers $a$ and $b$ satisfying the equality $3^a - 5^b = 2$.

(I. Gorodnin)
To find all integers \(a\) and \(b\) satisfying the equation \(3^a - 5^b = 2\), we will proceed with a detailed step-by-step analysis.

1. **Initial Check for Small Values:**
   - For \(a = 1\):
     \[
     3^1 - 5^b = 2 \implies 3 - 5^b = 2 \implies 5^b = 1 \implies b = 0
     \]
     Thus, \((a, b) = (1, 0)\) is a solution.
   - For \(a = 2\):
     \[
     3^2 - 5^b = 2 \implies 9 - 5^b = 2 \implies 5^b = 7
     \]
     Since \(5^b = 7\) has no integer solution for \(b\), there are no solutions for \(a = 2\).

2. **General Case Analysis:**
   - Rearrange the equation:
     \[
     3^a = 5^b + 2
     \]
   - Consider the equation modulo 3:
     \[
     3^a \equiv 0 \pmod{3} \implies 5^b + 2 \equiv 0 \pmod{3} \implies 5^b \equiv 1 \pmod{3}
     \]
     Since \(5 \equiv 2 \pmod{3}\), we have:
     \[
     2^b \equiv 1 \pmod{3}
     \]
     The powers of 2 modulo 3 cycle as \(2, 1\). Therefore, \(b\) must be even.

3. **Check for Small Even Values of \(b\):**
   - For \(b = 2\):
     \[
     3^a = 5^2 + 2 = 25 + 2 = 27 \implies 3^a = 27 \implies a = 3
     \]
     Thus, \((a, b) = (3, 2)\) is a solution.

4. **Verification of Solutions:**
   - For \((a, b) = (1, 0)\):
     \[
     3^1 - 5^0 = 3 - 1 = 2
     \]
     This is correct.
   - For \((a, b) = (3, 2)\):
     \[
     3^3 - 5^2 = 27 - 25 = 2
     \]
     This is also correct.

5. **Conclusion:**
   - We have found that the only solutions are \((a, b) = (1, 0)\) and \((a, b) = (3, 2)\).

The final answer is \( \boxed{ (1, 0) } \) and \((3, 2)\).