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 $k$?
 A: The formula quoted in Ribenboim has the general form,
$$u^2+dv^2 = (p^3 - 3 d p q^2)^2 + d(3 p^2 q - d q^3)^2 = (p^2+dq^2)^3\tag1$$
Assume $\gcd(u,v)=1$. For $d=3$, apparently it is integrally complete. 
But we cannot necessarily extend the same conclusion to general $d$. For example, for $d=11$, the formula above becomes,
$$(p^3 - 33 p q^2)^2 + 11(3 p^2 q - 11 q^3)^2 = (\color{blue}{p^2+11q^2})^3\tag2$$
The rational substitution $p,\,q = \frac{x+4y}{2},\;\frac{x}{2}$ transforms $(2)$ to,
$$(-4 x^3 - 15 x^2 y + 6 x y^2 + 8 y^3)^2 + 11(-x^3 + 3 x^2 y + 6 x y^2)^2 = (\color{blue}{3 x^2 + 2 x y + 4 y^2})^3\tag3$$
As W. Jagy pointed out, for $d=11$, the forms $x^2+11y^2,\;3 x^2 + 2 x y + 4 y^2$ are the desired sums. But for $p$ to be an integer, then $x$ has to be even which results in $\gcd(u,v)\neq1$. For $d=11$, then $(1)$ is only rationally complete.
For other $d$, say $d=47$, Pepin gave,
$$(13p^3 + 30p^2q - 42p q^2 - 18q^3)^2 + 47(p^3 - 6p^2q - 6p q^2 + 2q^3)^2 = 
  2^3(3p^2 + p q + 4q^2)^3\tag4$$
But now there is no rational substitution to transform $(1)$ to $(4)$. Thus, for certain $d$, then $(1)$ is not even complete rationally. (See also related posts: this and this.)
$\color{green}{Update:}$
As requested by OP, for $d=-6$, then formula $(1)$ yields,
$$(p^3 + 18 p q^2)^2 - 6(-3 p^2 q - 6 q^3)^2 = (p^2 - 6 q^2)^3\tag5$$
while this other formula, using initial $a,b,c = 5,\,2,\,1$ yields,
$$(5 r^3 - 36 r^2 s + 90 r s^2 - 72 s^3)^2 - 
  6(2 r^3 - 15 r^2 s + 36 r s^2 - 30 s^3)^2 = (r^2-6 s^2)^3\tag6$$
and there is no rational transformation between $(5)$ and $(6)$. Thus, for $d=-6$, then $(1)$ is not rationally complete also.
A: 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.
$$x=p^6+q(b^2+8bs-5s^2)p^4+q^2(s^2-b^2)(b^2-8bs-5s^2)p^2+q^3(s^2-b^3)^3$$
$$y=2p(q^2(2s+b)(s^2-b^2)^2+2qb(b^2-3s^2)p^2-(2s-b)p^4)$$
$$z=p^4+2q(s^2+b^2)p^2+q^2(s^2-b^2)^2$$
