= Solution
Use the scalar components of the three <chiral superfields>, and set $a=|\phi_0|^2$, $b=|\phi_+|^2$, $c=|\phi_-|^2$ and $\Sigma=a+b+c$. We assume the physical sign $g>0$ and a nonzero coupling $\lambda$. The coefficient $g$ below is exactly the coefficient in the supplied <D-term scalar potential>; it should not be silently replaced by a differently normalized squared <gauge coupling>. The canonical <Kähler potential> has identity <Kähler metric>, and the three <Kähler covariant derivatives of a superpotential> are
$$
\begin{aligned}
D_0W&=\lambda\phi_+\phi_-(1+\kappa^2a),\\
D_+W&=\lambda\phi_0\phi_-(1+\kappa^2b),\\
D_-W&=\lambda\phi_0\phi_+(1+\kappa^2c).
\end{aligned}
$$
Substitution in the <supergravity F-term potential> gives
$$
\begin{aligned}
V_F={}&|\lambda|^2e^{\kappa^2\Sigma}
\left[bc(1+\kappa^2a)^2+ac(1+\kappa^2b)^2
+ab(1+\kappa^2c)^2-3\kappa^2abc\right]\\
={}&|\lambda|^2e^{\kappa^2\Sigma}
\left[ab+ac+bc+3\kappa^2abc+\kappa^4abc\Sigma\right].
\end{aligned}
$$
In particular, the subtraction cancels only half of the six $\kappa^2abc$ terms contributed by the three squares. The full <scalar potential> is
$$
\boxed{V=|\lambda|^2e^{\kappa^2\Sigma}
\left[ab+ac+bc+3\kappa^2abc+\kappa^4abc\Sigma\right]
+g(b-c-\zeta)^2\ge0.}
$$
Its nonnegativity follows term by term from $a,b,c\ge0$ and $g>0$. This is a special property of this cubic <superpotential>, not of a general <supergravity F-term potential>.
The <D-term vacuum branches of a cubic superpotential> must be described as branches: there are not two isolated field-space minima. First suppose $\zeta>0$. Vanishing of $V_F$ requires $ab=ac=bc=0$, and vanishing of the <D-term scalar potential> requires $b-c=\zeta$. Together these conditions give the <vacuum expectation values>
$$
\boxed{\langle\phi_0\rangle=\langle\phi_-\rangle=0,
\qquad |\langle\phi_+\rangle|=\sqrt\zeta,
\qquad V=0.}
$$
The phase of $\phi_+$ lies on a <gauge orbit>, and one may choose it positive and real. At this <global minimum>, all $D_iW$ and the real <auxiliary field> of the <vector multiplet> vanish. It is a <supersymmetric vacuum>, even though the internal <gauge symmetry> is spontaneously broken. For $\zeta<0$, interchange $\phi_+$ and $\phi_-$: the nonzero <vacuum expectation value> has magnitude $\sqrt{-\zeta}$. For $\zeta=0$, the <supersymmetric vacuum> is the entire family $\phi_+=\phi_-=0$, with $\phi_0$ arbitrary.
The other branch has $\phi_+=\phi_-=0$ and arbitrary $\phi_0$. Here $W=D_iW=0$ but the real <auxiliary field> is nonzero for $\zeta\ne0$, and
$$
V=g\zeta^2.
$$
Thus this branch has pure <D-term> <supersymmetry breaking> and a <flat direction of a scalar potential> in the neutral scalar. It is not stable at every neutral-field value. At a fixed $a=|\phi_0|^2$, expansion in the charged scalars yields
$$
V=g\zeta^2
+\left(|\lambda|^2ae^{\kappa^2a}-2g\zeta\right)|\phi_+|^2
+\left(|\lambda|^2ae^{\kappa^2a}+2g\zeta\right)|\phi_-|^2
+O\left((b+c)^2\right).
$$
Hence the two charged squared <masses> and the stability condition are
$$
\boxed{m_\pm^2=|\lambda|^2ae^{\kappa^2a}\mp2g\zeta,
\qquad |\lambda|^2ae^{\kappa^2a}>2g|\zeta|.}
$$
Above this threshold, both charged directions have positive <mass> squared. Continuity makes them uniformly positive in a sufficiently small neighborhood of the chosen neutral-field value, so the stationary branch is a genuine, non-strict <local minimum> valley. Its neutral direction stays flat. Below the threshold, a charged <mass> squared is negative and the branch is a <saddle point>. At equality it is a boundary of the stable valley, not a <local minimum> in the full field space: changing $a$ slightly downward and switching on the unstable charged field lowers $V$. For example, for $\zeta>0$, put $a=a_c-\epsilon$, $b=\epsilon^2$, $c=0$. Since $M(a)=|\lambda|^2ae^{\kappa^2a}$ has $M'(a_c)>0$, one finds $V-g\zeta^2=-M'(a_c)\epsilon^3+O(\epsilon^4)<0$.
There are no further local-minimum branches under these generic assumptions. If $b+c>0$, the expression for $V_F$ is strictly increasing in $a$, so a stationary configuration with a charged condensate must have $a=0$. At $a=0$, an interior critical point with $b,c>0$ is impossible, because
$$
\partial_b V+\partial_c V
=|\lambda|^2e^{\kappa^2(b+c)}
\left(b+c+2\kappa^2bc\right)>0.
$$
On the boundary $c=0$, the relevant <scalar potential> is $g(b-\zeta)^2$; on $b=0$ it is $g(c+\zeta)^2$. Their nonnegative-domain minima give exactly the <supersymmetric vacuum> described above, or the origin of the charged-field branch. Thus, for nonzero $\zeta$, \b[the two types are the zero-energy supersymmetric gauge orbit and the positive-energy D-breaking minimum valley above the threshold]. If $\zeta=0$, the two types coalesce into the flat <supersymmetric vacuum>; the literal two-minimum assertion needs this parameter qualification.
In the global <supersymmetry> limit $\kappa\to0$, the <F-term scalar potential> becomes $|\lambda|^2(ab+ac+bc)$ and the <D-term scalar potential> retains its form. The same two types of branches occur, with the threshold $|\lambda|^2a>2g|\zeta|$. <Supergravity> instead introduces the exponential and additional positive terms. The <goldstino> of global <supersymmetry breaking> is a physical <fermion>; in local <supersymmetry> it participates in the <super-Higgs mechanism>. A positive vacuum energy also gravitates in <supergravity>. In particular, the positive-energy branch here has $W=0$ and hence vanishing <gravitino mass from a superpotential>; it is not the tuned zero-energy broken vacuum of Question 3.
There is a further gauge-consistency qualification. In conventional two-derivative matter-coupled <supergravity>, a constant <Fayet–Iliopoulos term> gauges an <R-symmetry> and requires a corresponding nontrivial transformation of the <superpotential>. The apparent charges $0,+1,-1$ make the supplied cubic <superpotential> neutral, so a nonzero independent constant $\zeta$ does not by itself specify a consistent conventional <supergravity> action. The computed branches are those of the supplied <scalar potential>; a full embedding needs compatible charges or additional fields. This restriction is explained in https://arxiv.org/abs/hep-th/0410053[Van Proeyen's derivation of the superpotential gauge-covariance condition].
More generally, the <supergravity moment-map constraint on D-term breaking> follows directly from gauge covariance. If $k^i\partial_iW=-\kappa^2rW$ and the <moment map> is $\mathcal P=i(k^iK_i-r)$, then at a point with $W\ne0$,
$$
\mathcal P=i\kappa^{-2}\frac{k^iD_iW}{W}.
$$
Consequently all chiral <auxiliary fields> vanishing implies the <D-term> also vanishes there. This relation has no analogue forcing that implication in an arbitrary global <supersymmetry> model. Its $W\ne0$ hypothesis matters: the formal D-breaking branch above has $W=0$, so it does not contradict the identity. These conventional gauge-completion restrictions are additional to the elementary minimization calculation.
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