I follow Kelly [Basic concepts of enriched category theory].
Q1. In general, no; in practice, yes.
By definition, we have an equaliser diagram
$$\int_{i : \mathcal{I}} \mathcal{A} (F (i), G (i)) \rightarrow \prod_{i \in \operatorname{ob} \mathcal{I}} \mathcal{A} (F (i), G (i)) \rightrightarrows \prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \mathcal{V} (\mathcal{I} (i, j), \mathcal{A} (F (i), G (j)))$$
and applying the underlying set functor preserves limits, so we get an equaliser diagram of sets:
$$\left( \int_{i : \mathcal{I}} \mathcal{A} (F (i), G (i)) \right)_0 \rightarrow \prod_{i \in \operatorname{ob} \mathcal{I}} \mathcal{A}_0 (F (i), G (i)) \rightrightarrows \prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \mathcal{V}_0 (\mathcal{I} (i, j), \mathcal{A} (F (i), G (j)))$$
This is almost – but not quite – the definition of $\int_{i : \mathcal{I}_0} \mathcal{A}_0 (F (i), G (i))$.
For the latter we would have $\prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \textbf{Set} (\mathcal{I}_0 (i, j), \mathcal{A}_0 (F (i), G (j)))$ instead.
For every $V$ and $W$ in $\mathcal{V}$, the action of the underlying set functor $\mathcal{V}_0 \to \textbf{Set}$ is a map
$$\mathcal{V}_0 (V, W) \longrightarrow \textbf{Set} (V_0, W_0)$$
so we get a canonical map:
$$\prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \mathcal{V}_0 (\mathcal{I} (i, j), \mathcal{A} (F (i), G (j))) \longrightarrow \prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \textbf{Set} (\mathcal{I}_0 (i, j), \mathcal{A}_0 (F (i), G (j))) \tag{†}$$
This induces a comparison map
$$\left( \int_{i : \mathcal{I}} \mathcal{A} (F (i), G (i)) \right)_0 \longrightarrow \int_{i : \mathcal{I}_0} \mathcal{A}_0 (F (i), G (i)) \tag{‡}$$
and it is easy to see that (‡) is a bijection if (†) is an injection.
This happens if the underlying set functor $\mathcal{V}_0 \to \textbf{Set}$ is faithful, which is usually the case in practice.
(The main exception is when $\mathcal{V}$ is a category of multisorted structures, e.g. simplicial sets or chain complexes.)
Q2.
Yes.
When $V$ is a coproduct of $X$ copies of the monoidal unit, then the underlying set functor gives a bijection $\mathcal{V}_0 (V, W) \to \textbf{Set} (X, W)$.
Thus, if $\mathcal{I}$ is the free $\mathcal{V}$-category on an ordinary category, then (†) is a bijection, so (‡) is also a bijection.
(But we could have deduced this directly from the universal property of free $\mathcal{V}$-categories!)
Q3.
Yes.
Suppose $F$ and $G$ are constant functors, with values $A$ and $B$ respectively.
You can directly verify that the diagonal morphism $\mathcal{A} (A, B) \to \prod_{i \in \operatorname{ob} \mathcal{I}} \mathcal{A} (F (i), G (i))$ equalises $\prod_{i \in \operatorname{ob} \mathcal{I}} \mathcal{A} (F (i), G (i)) \rightrightarrows \prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \mathcal{V} (\mathcal{I} (i, j), \mathcal{A} (F (i), G (j)))$ and so we have the desired morphism $\mathcal{A} (A, B) \to \int_{i : \mathcal{I}} \mathcal{A} (F (i), G (i))$.
(Again, you could instead use the universal property of free $\mathcal{V}$-categories here.
The point is that a $\mathcal{V}$-functor $\mathcal{A} \to [\mathcal{I}, \mathcal{A}]$ is essentially the same thing as a $\mathcal{V}$-functor $\mathcal{I} \to [\mathcal{A}, \mathcal{A}]$, and when $\mathcal{I}$ is free we can just take the constant functor with value $\textrm{id}_\mathcal{A}$.)
Best Answer
Kelly doesn't explicitly speak about enriched terminal objects in his book, but about weighted limits and the special case of conical limits. If define a terminal object $1$ to be a limit over the empty diagram (in which case there is a unique weight), you get the condition that all hom objects $\mathrm{hom}(A,1)$ are terminal in $\mathcal{V}$, as you suspected.