- Yes, but perhaps in an other sense than you may think.
- I have nothing to say :-)
- Yes.
- Yes.
$\newcommand{\mor}[3]{#1 \colon #2 \rightarrow #3}%
\newcommand{\catl}[1]{\mathbb{#1}}%
\newcommand{\catw}[1] {\mathbf{#1}}$

Here is my elaboration.

Let me first omit some irrelevant details. We shall say that a functor $\mor{F}{\catl{C}}{\catl{D}}$ has a left multiadjoint if for every $X \in \catl{D}$, the hom-functor $\hom(X, F(-))$ is a (small) coproduct of representables:
$$\hom(X, F(-)) \approx \coprod_i\hom(G_i(X), -)$$
At this point I am not sure yet if I fully understand the above definition --- it makes me wonder if there is anything so special about coproducts: i.e. what if we substituted copoducts with other classes of colimits? Let me try:

A functor $\mor{F}{\catl{C}}{\catl{D}}$ has a left adjoint if for every $X \in \catl{D}$, the hom-functor $\hom(X, F(-))$ is representable:
$$\hom(X, F(-)) \approx \hom(G(X), -)$$
I have nothing to add here.

A functor $\mor{F}{\catl{C}}{\catl{D}}$ has a left nothingadjoint if for every $X \in \catl{D}$ the hom-functor $\hom(X, F(-))$, is a (small) colimit of representables:
$$\hom(X, F(-)) \approx \mathit{colim}_i\hom(G_i(X), -)$$
Nothingadjointness is not an interesting concept, because by Yoneda every $\catw{Set}$-valued functor is a colimit of representables.

A functor $\mor{F}{\catl{C}}{\catl{D}}$ has a finitary approximation to a left adjoint if for every $X \in \catl{D}$, the hom-functor $\hom(X, F(-))$ is a *filtered* colimit of representables:
$$\hom(X, F(-)) \approx \mathit{fcolim}_i\hom(G_i(X), -)$$
In the literature, a functor that has a finitary approximation to a left adjoint is called (left) *flat*.

Another way of looking at the above three situations is that we want to "represent" objects (apologize for contravariance, but I have not realized that I am describing a contravariant world until now) from $\catw{Set}^\catl{C}$ in $\catl{C}^{op}$, in $\catw{Set}^\catl{C}$ (i.e. a free cocompletion of $\catl{C}^{op}$) and in $\mathit{Ind}(\catl{C}^{op})$ (i.e. a free cofiltered completion of $\catl{C}^{op}$) respectively. A reasonable person, who wants to better understand multiadjunctions, should start looking now for free coproduct completion of a category.

Let $\catl{B}$ be a locally small category. One may associate with it the canonical $\catw{Set}$-indexing functor $\mor{\mathit{fam}(\catl{B})}{\catw{Set}^{op}}{\catw{Cat}}$:
$$X \mapsto \catl{B}^X$$
One may think of $\catl{B}^X$ as of the category of formal $X$-indexed coproducts of objects from $\catl{B}$. The Grothendieck construction for $\mathit{fam}(\catl{B})$ glues categories $\catl{B}^X$ of formal $X$-indexed coproducts along sets $X$, giving the category:
$$\int \mathit{fam}(\catl{B})$$
which is a formal (small) coproduct completion of $\catl{B}$ (of course, one needs to carefully check this statement). In fact, the above construction rises to a monad on $\catw{Cat}$, and one may develop a formal theory of multiadjunctions (similar to the formal theory of adjunctions through distributors) in the 2-category of Kleisly resolution of the monad.

Nonetheless, there is a less heavy explanation. I claim that a functor $\mor{F}{\catl{C}}{\catl{D}}$ has a left multiadjoint if $\mor{{F^{op}}^\star}{\int \mathit{fam}(\catl{C}^{op})}{\int \mathit{fam}(\catl{D}^{op})}$ defined as:
$${F^{op}}^\star(\{C_i\}_{i \in I}) = \{F^{op}(C_i)\}_{i \in I}$$
has right adjoint. First, let me show the trivial direction --- assume that ${F^{op}}^\star$ has right adjoint. In particular, this gives us:
$$\hom_{\int \mathit{fam}(\catl{D}^{op})}({F^{op}}^\star(\{C_i\}_{i \in 1}), \{D_j\}_{j \in 1}) \approx \hom_{\int \mathit{fam}(\catl{C}^{op})}(\{C_i\}_{i \in 1}, G(\{D_j\}_{j \in 1}))$$
which simplifies to:
$$\hom_{\int \mathit{fam}(\catl{D}^{op})}(F^{op}(C), \{D\}) \approx \hom_{\int \mathit{fam}(\catl{C}^{op})}(\{C\}, G(\{D\}))$$
A morphism $F^{op}(C) \rightarrow \{D\}$ in $\int \mathit{fam}(\catl{D}^{op})$ is just a morphism $D \rightarrow F(C)$ in $\catl{D}$. Similarly, a single morphism $\{C\} \rightarrow G(\{D\})$ in $\int \mathit{fam}(\catl{C}^{op})$ is a morphism $G(\{D\})_k \rightarrow C$ in $\catl{C}$, for one $k \in K$, where $K$ is the indexing set of $G(\{D\})$. So:
$$\hom_{\int \mathit{fam}(\catl{C}^{op})}(\{C\}, G(\{D\}))\approx \coprod_{k\in K} \hom_\catl{C}(G(\{D\})_k, C)$$
and we get the formula for left multiadjunction:
$$\hom_\catl{D}(F(C), D) \approx \coprod_{k\in K} \hom_\catl{C}(G(\{D\})_k, C)$$
In the other direction, let us assume that the above formula holds, and freely extend $G$ to ${G^{op}}^\star$:
$${G^{op}}^\star(\{D_j\}_{j \in J}) = \coprod_{j\in J} G(\{D_j\})$$
Since ${F^{op}}^\star$ is free, it suffices to show:
$$\hom_{\int \mathit{fam}(\catl{D}^{op})}({F^{op}}^\star(\{C\}), \{D_j\}_{j \in J}) \approx \hom_{\int \mathit{fam}(\catl{C}^{op})}(\{C\}, {G^{op}}^\star(\{D_j\}_{j \in J}))$$
The left side is isomorphic to:
$$\coprod_{j \in J} \hom_\catl{D}(D_j, F(C))$$
whereas, the right side is isomorphic to:
$$\coprod_{j \in J} \coprod_{k \in K} \hom_\catl{C}(G(\{D_j\})_k, C)$$
Thus ${G^{op}}^\star$ is right adjoint to ${F^{op}}^\star$.

Moving back to your questions:

- Yes, they are colimits in the free coproduct completion of a category.
- Still nothing to say :-)
- Yes, because it is generally true for (co)limits.
- Yes, because the coproduct completion, being a 2-functor, preserves adjunctions.