Solution 1: If we subtract the two equations, we obtain
$$ax^2 + 8 - 8x^2 - a = (a-8)(x^2-1) = (a-8)(x+1)(x-1) = 0,$$
thus either $a = 8$ or $x = -1$ or $x = 1$. If $a = 8$, we are left with
$$5x^3 + 8x^2 + 8 = (x+2)(5x^2 - 2x + 4) = 0.$$
The second factor has no real roots, since its discriminant $2^2 - 4 \cdot 5 \cdot 4 = -76$ is negative.
Thus $x = -2$ in this case.
If $x = -1$, we get $a = -5x^3 - 8x^2 = -3$, and if $x = 1$, we get $a = -5x^3 - 8x^2 = -13$.
In summary, there are three possible pairs: $(a, x) = (8, -2)$, $(a, x) = (-3, -1)$ and $(a, x) = (-13, 1)$.

Solution 2: Solving the second equation for $a$ gives us $a = -5x^3 - 8x^2$, thus
$$5x^3 + (-5x^3 - 8x^2)x^2 + 8 = -5x^5 - 8x^4 + 5x^3 + 8 = 0.$$
The polynomial can be factorised:
$$\begin{aligned} -5x^5 - 8x^4 + 5x^3 + 8 &= 5x^3(1 - x^2) + 8(1 - x^4) = 5x^3(1 - x^2) + 8(1 + x^2)(1 - x^2) \\ &= (1 - x^2)(5x^3 + 8x^2 + 8) = (1 - x)(1 + x)(x + 2)(5x^2 - 2x + 4). \end{aligned}$$
Again, we find that $x = 1$, $x = -1$ or $x = -2$, since the quadratic factor has no real roots.
The value of $a$ is obtained from the equation $a = -5x^3 - 8x^2$, so we end up with the three possibilities $(a, x) = (8, -2)$, $(a, x) = (-3, -1)$ and $(a, x) = (-13, 1)$ again.