Find all pairs $(p,q)$ of positive primes such that the equation $3x^2 - px + q = 0$ has two distinct rational roots.
To find all pairs \((p, q)\) of positive primes such that the equation \(3x^2 - px + q = 0\) has two distinct rational roots, we can follow these steps:

1. **Apply the Rational Root Theorem**:
   According to the Rational Root Theorem, if \(\frac{a}{b}\) is a rational root of the polynomial \(3x^2 - px + q = 0\), then \(a\) must be a divisor of the constant term \(q\) and \(b\) must be a divisor of the leading coefficient \(3\). Therefore, the possible values for \(a\) are \(\pm 1, \pm q\) and for \(b\) are \(\pm 1, \pm 3\).

2. **List all possible rational roots**:
   The possible rational roots are:
   \[
   \pm 1, \pm q, \pm \frac{1}{3}, \pm \frac{q}{3}
   \]

3. **Consider the distinct roots condition**:
   Since the roots must be distinct and rational, we need to consider the combinations of these roots. We also need to ensure that the polynomial remains a quadratic with integer coefficients.

4. **Analyze the combinations**:
   - If both roots have \(b = 3\), the leading coefficient would be \(9\), which is not possible since the leading coefficient is \(3\).
   - If both roots have \(b = 1\), the leading coefficient remains \(3\), but this would imply that \(p\) and \(q\) are not primes unless the polynomial is \(3(x^2 - x + 1)\), which does not work.

5. **Consider one root with \(b = 3\) and the other with \(b = 1\)**:
   - Let one root be \(\frac{a}{3}\) and the other be \(b\), where \(a\) is a divisor of \(q\) and \(b\) is a divisor of \(3\).

6. **Form the polynomial with these roots**:
   - Suppose the roots are \(\frac{1}{3}\) and \(q\). Then the polynomial can be written as:
     \[
     3 \left( x - \frac{1}{3} \right) \left( x - q \right) = 3 \left( x^2 - \left( q + \frac{1}{3} \right) x + \frac{q}{3} \right)
     \]
     Simplifying, we get:
     \[
     3x^2 - \left( 3q + 1 \right) x + q = 0
     \]
     This implies \(p = 3q + 1\).

   - Suppose the roots are \(\frac{q}{3}\) and \(1\). Then the polynomial can be written as:
     \[
     3 \left( x - \frac{q}{3} \right) \left( x - 1 \right) = 3 \left( x^2 - \left( 1 + \frac{q}{3} \right) x + \frac{q}{3} \right)
     \]
     Simplifying, we get:
     \[
     3x^2 - \left( q + 3 \right) x + q = 0
     \]
     This implies \(p = q + 3\).

7. **Check for prime pairs**:
   - For \(p = 3q + 1\):
     - If \(q = 2\), then \(p = 3(2) + 1 = 7\). Both \(2\) and \(7\) are primes.
   - For \(p = q + 3\):
     - If \(q = 2\), then \(p = 2 + 3 = 5\). Both \(2\) and \(5\) are primes.

Thus, the only pairs \((p, q)\) that satisfy the conditions are \((5, 2)\) and \((7, 2)\).

The final answer is \(\boxed{(p, q) = (5, 2), (7, 2)}\)