Three questions about the form $X^2 \pm 3Y^2 = Z^3$ and a related lemma In Ribenboim’s Fermat’s Last Theorem for Amateurs, he gives the following lemma [Lemma 4.7, pp. 30–31].
Lemma. Let $E$ be the set of all triples $(u, v, s)$ such that $s$ is odd, $\gcd(u,v) = 1$ and $s^3 = u^2 + 3v^2$. Let $F$ be the set of all pairs $(t,w)$ where $\gcd(t, w) = 1$ and $t \not\equiv w\!\pmod{2}$. The mapping $\Phi  : F \rightarrow E$ given by $\Phi (t, w) = (u, v, s)$ with
\begin{cases}
\, u = t(t^2−9w^2),  \\
\, v = 3w(t^2-w^2),\\
\, s = t^2 + 3w^2,
\end{cases}
is onto $E$.
Q1: Am I understanding “onto” correctly by interpreting this to mean that this is a complete integer parameterization of the form $X^2 + 3Y^2 = Z^3$?
Q2: Is there a similar solution for the form $X^2-3Y^2 = Z^3$? It appears that
\begin{cases}
\, x = t(t^2+9w^2),  \\
\, y = 3w(t^2+w^2),\\
\, z = t^2 - 3w^2,
\end{cases}
satisfies, but I want to be sure it’s complete.
Q3: Is this generalizable to the form $X^2 \pm kY^2 = Z^3$ for any [possibly non-square or squarefree] integer $k$?

To describe the solutions of the equation.  $$x^2+qy^2=z^3$$
I think best would be to describe a solution using $3$ parameters.
$$x=p^6+q(b^2+8bs-5s^2)p^4+q^2(s^2-b^2)(b^2-8bs-5s^2)p^2+q^3(s^2-b^3)^3$$
$$y=2p(q^2(2s+b)(s^2-b^2)^2+2qb(b^2-3s^2)p^2-(2s-b)p^4)$$
$$z=p^4+2q(s^2+b^2)p^2+q^2(s^2-b^2)^2$$
