Find all pairs of positive integers $(x, y)$ that satisfy the following two conditions:
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[*] The average of $x$ and $y$ is a prime.
[*] $\frac{x!+y!}{x+y}$ is an integer.
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1. **Assume without loss of generality (WLOG) that \( x \geq y \).** This assumption simplifies our analysis due to the symmetry of the problem.

2. **Let the average of \( x \) and \( y \) be a prime number \( p \).** This gives us the equation:
   \[
   \frac{x + y}{2} = p \implies x + y = 2p
   \]
   Since \( p \) is a prime number, \( x \) and \( y \) must be such that their sum is twice a prime number.

3. **Given that \( \frac{x! + y!}{x + y} \) is an integer, substitute \( x + y = 2p \):**
   \[
   \frac{x! + y!}{2p} \text{ is an integer}
   \]

4. **Since \( x \geq y \), we have \( x \geq p \).** This implies that \( p \) divides \( x! \) because \( x! \) contains all integers from 1 to \( x \), including \( p \).

5. **For \( \frac{x! + y!}{2p} \) to be an integer, \( p \) must also divide \( y! \).** Since \( y \leq x \) and \( x \geq p \), we have \( y \leq p \).

6. **Combining the inequalities \( x \geq p \) and \( y \leq p \) with the equation \( x + y = 2p \), we get:**
   \[
   x = y = p
   \]
   This is because if \( x \) were greater than \( p \), then \( y \) would have to be less than \( p \) to maintain the sum \( 2p \), but this would contradict the requirement that \( p \) divides \( y! \) unless \( y = p \).

7. **Thus, the only solution is \( x = y = p \) for any prime \( p \).**

The final answer is \( \boxed{ (p, p) = (x, y) } \) for any prime \( p \).