a.
The roots of $z^2 + 3z + 8 = 0$ are $\frac{-3 \pm \sqrt{23}i}{2}$. Let $\alpha = \frac{-3 + \sqrt{23}i}{2}$. Then $\bar{\alpha} = \frac{-3 - \sqrt{23}i}{2}$, and hence $x^2 + 3xy + 8y^2 = (x - \alpha y)(x - \bar{\alpha}y)$. Note that
$$\begin{aligned} (x_1 - \alpha y_1)(x_2 - \alpha y_2) &= (x_1 x_2 + \alpha^2 y_1 y_2) - \alpha(x_1 y_2 + x_2 y_1) \\ &= (x_1 x_2 + (-3\alpha - 8)y_1 y_2) - \alpha(x_1 y_2 + x_2 y_1) \\ &= (x_1 x_2 - 8y_1 y_2) - \alpha(x_1 y_2 + x_2 y_1 + 3y_1 y_2). \end{aligned}$$
Thus, defining $s = x_1x_2 - 8y_1y_2$ and $t = x_1y_2 + x_2y_1 + 3y_1y_2$, we have
$$\begin{aligned} & (x_1^2 + 3x_1y_1 + 8y_1^2)(x_2^2 + 3x_2y_2 + 8y_2^2) \\ &= (x_1 - \alpha y_1)(x_1 - \bar{\alpha} y_1)(x_2 - \alpha y_2)(x_2 - \bar{\alpha} y_2) \\ &= (x_1 - \alpha y_1)(x_2 - \alpha y_2)(\overline{(x_1 - \alpha y_1)(x_2 - \alpha y_2)}) \\ &= (s - \alpha t)\overline{(s - \alpha t)} \\ &= (s - \alpha t)(s - \bar{\alpha} t) \\ &= s^2 + 3st + 8t^2. \end{aligned}$$
This clearly proves the result.

b.
No. Suppose on the contrary that $x^2 + 3xy + 8y^2 \equiv 7 \pmod{23}$ for some integers $x$ and $y$. This implies $4x^2 + 12xy + 32y^2 \equiv 28 \pmod{23}$, and hence $(2x + 3y)^2 \equiv 5 \pmod{23}$. However, we can check that 5 is not a square modulo 23 by computing the Legendre symbol $\binom{5}{23} = \binom{23}{5} = \binom{3}{5} = -1$ (or by testing $0^2, (\pm 1)^2, \dots, (\pm 11)^2 \pmod{23}$). This is a contradiction, and so there is no such integer in $S$.