a. Let $x + \sqrt{3} = q$, $x^2 + \sqrt{3} = p$ with $p, q \in \mathbb{Q}$. Then
$$x = q - \sqrt{3} \Rightarrow x^2 = q^2 - 2q\sqrt{3} + 3$$
so substituting in the second one gives:
$$(q^2 - 2q\sqrt{3} + 3) + \sqrt{3} = p \Leftrightarrow -\sqrt{3}(2q-1) = p - q^2 - 3$$
It follows that $2q-1=0 \Leftrightarrow q = \frac{1}{2}$. In this case, $p = q^2+3 \Rightarrow p = \frac{1}{4} + 3 = \frac{13}{4}$ and
$x = \frac{1}{2} - \sqrt{3}$.

b. Let $y + \sqrt{3} = q$, $y^3 + \sqrt{3} = p$ with $p, q \in \mathbb{Q}$. Then
$$y = q - \sqrt{3} \Rightarrow y^3 = q^3 - 3q^2\sqrt{3} + 9q - 3\sqrt{3}$$
so substituting in the second one gives:
$$(q^3 - 3q^2\sqrt{3} + 9q - 3\sqrt{3}) + \sqrt{3} = p \Leftrightarrow -\sqrt{3}(3q^2 + 2) = p - q^3 - 9q \Leftrightarrow \\ \sqrt{3} = \frac{q^3 + 9q - p}{3q^2 + 2} \in \mathbb{Q}$$
which is absurd.