Determine all pairs of real numbers $a$ and $x$ that satisfy the simultaneous equations $$5x^3 + ax^2 + 8 = 0$$ and $$5x^3 + 8x^2 + a = 0.$$
1. Given the simultaneous equations:
   \[
   5x^3 + ax^2 + 8 = 0
   \]
   and
   \[
   5x^3 + 8x^2 + a = 0
   \]

2. Subtract the second equation from the first:
   \[
   (5x^3 + ax^2 + 8) - (5x^3 + 8x^2 + a) = 0
   \]
   Simplifying, we get:
   \[
   ax^2 + 8 - 8x^2 - a = 0
   \]
   \[
   (a - 8)x^2 + 8 - a = 0
   \]

3. Rearrange the terms:
   \[
   (a - 8)x^2 + (8 - a) = 0
   \]
   Factor out \((a - 8)\):
   \[
   (a - 8)(x^2 - 1) = 0
   \]

4. This gives us two cases:
   \[
   a - 8 = 0 \quad \text{or} \quad x^2 - 1 = 0
   \]
   Therefore:
   \[
   a = 8 \quad \text{or} \quad x = \pm 1
   \]

5. Case 1: \(x = 1\)
   Substitute \(x = 1\) into the first equation:
   \[
   5(1)^3 + a(1)^2 + 8 = 0
   \]
   \[
   5 + a + 8 = 0
   \]
   \[
   a + 13 = 0
   \]
   \[
   a = -13
   \]
   Thus, one solution pair is \((-13, 1)\).

6. Case 2: \(x = -1\)
   Substitute \(x = -1\) into the first equation:
   \[
   5(-1)^3 + a(-1)^2 + 8 = 0
   \]
   \[
   -5 + a + 8 = 0
   \]
   \[
   a + 3 = 0
   \]
   \[
   a = -3
   \]
   Thus, another solution pair is \((-3, -1)\).

7. Case 3: \(a = 8\)
   Substitute \(a = 8\) into the first equation:
   \[
   5x^3 + 8x^2 + 8 = 0
   \]
   Factor the equation:
   \[
   5x^3 + 8x^2 + 8 = 0
   \]
   \[
   (x + 2)(5x^2 - 2x + 4) = 0
   \]
   This gives:
   \[
   x = -2 \quad \text{or} \quad 5x^2 - 2x + 4 = 0
   \]
   The quadratic equation \(5x^2 - 2x + 4 = 0\) has no real solutions (discriminant is negative).

   Thus, the only solution for \(a = 8\) is \(x = -2\), giving the solution pair \((8, -2)\).

8. Combining all the solutions, we have:
   \[
   (-13, 1), (-3, -1), (8, -2)
   \]

The final answer is \(\boxed{(-13, 1), (-3, -1), (8, -2)}\)