I will try to answer questions 1, 2, and 4. Question 3 is somewhat orthogonal to them, and I don't know the answer.
Clearly (1) implies both (2) and (3).
Suppose that (3) holds. Let me write $\sigma$ for the symmetry on $\mathcal{C}$, and consider the map $coev' := \sigma \circ ev^R : I \to X \otimes X^\vee$ and $ev' := coev^R \circ \sigma : X^\vee \otimes X \to I$. The zorro equations for $coev, ev$ imply zorro equations for $coev', ev'$. In other words, $coev'$ and $ev'$ select another 1-duality between $X$ and $X^\vee$.
On the other hand, given $X$, the space of duality data $(X^\vee, coev, ev)$ is contractible [I will only use that it's connected] if it is nonempty. Thus, there is an automorphism of $X^\vee$ which conjugates $coev$ to $coev'$ and $ev$ to $ev'$. Of course, this automorphism is precisely the Serre, but I don't really care what its formula is, only that it exists. (Anyway, you can work out a formula: you just use the only formula that identifies different duals with each other.)
Isomorphisms are in particular adjunctible. In particular, since $coev'$ is $coev$ composed with some isomorphisms, they have the same amount of adjunctibility. But $coev$ is by assumption right-adjunctible, whereas $coev' = \sigma \circ ev^R$ is by construction left-adjunctible. So both $coev$ and $coev'$ are both-sides-adjunctible, and ditto for $ev$ and $ev'$.
Now you repeat the game. For example, $coev'$ and $ev'$ admit right adjoints $(coev')^R$ and $(ev')^R$, from which you can build yet another duality $coev'', ev''$ between $X,X^\vee$, and from it conclude that all the players in the game are twice-right-adjunctible and twice-left-adjunctible. Keep going to conclude that (3) implies (1).
To finish, I need to show that (2) implies (3). More generally, I want to show the following. Suppose I have a dual pair $(X,X^\vee, coev, ev)$, and I know that $ev : X^\vee \otimes X \to I$ admits a left adjoint $ev^L : I \to X^\vee \otimes X$. Then I claim that $coev : I \to X \otimes X^\vee$ admits a right adjoint $coev^R : X \otimes X^\vee \to I$.
There is only one possible guess for the formula for $coev^R$:
$$ coev^R := (ev \otimes ev) \circ (\sigma \otimes \sigma) \circ (X^\vee \otimes ev^L \otimes X).$$
[Notation: I am using the same name for an object and for its identity 1-morphism.]
The adjunction $ev^L \dashv ev$ consists of 2-morphisms $I \Rightarrow ev \circ ev^L$ and $ev^L \circ ev \Rightarrow X^\vee \otimes X$. I want to supply 2-morphisms $I \Rightarrow coev^R \circ coev$ and $coev \circ coev^R \Rightarrow X \otimes X^\vee$.
But the zorro relation for $ev,coev$ [and some interchanges] provides an isomorphism
$$coev^R \circ coev \cong ev \circ ev^L$$
and I already have a 2-morphism $I \Rightarrow ev \circ ev^L$, so just compose it with my zorro-provided isomorphism to get the desired $I \Rightarrow coev^R \circ coev$.
On the other hand,
$$ coev \cong (X \otimes ev \otimes X^\vee) \circ (coev \otimes coev),$$
and so
$$ coev \circ coev^R \cong [(X \otimes ev \otimes X^\vee) \circ (coev \otimes coev)] \circ [(ev \otimes ev) \circ (\sigma \otimes \sigma) \circ (X^\vee \otimes ev^L \otimes X)].$$
Now apply some interchanges to bring $ev, ev^L$ next to each other:
$$ \cong (X \otimes ev\sigma \otimes ev\sigma \otimes X^\vee) \circ (\sigma \otimes ev^Lev \otimes \sigma) \circ (X \otimes coev \otimes coev \otimes X^\vee)$$
But $ev^Lev$ emits a 2-morphism to $X^\vee \otimes X$, so my complicated composition emits a 2-morphism to
$$ \Rightarrow (X \otimes ev\sigma \otimes ev\sigma \otimes X^\vee) \circ (\sigma \otimes X^\vee \otimes X \otimes \sigma) \circ (X \otimes coev \otimes coev \otimes X^\vee),$$
which, by some zorro and interchange laws, is isomorphic to
$$ \cong X \otimes X^\vee.$$
The last step, which I'll leave to you to enjoy, is to check that indeed the 2-morphisms I just constructed do supply an adjunction $coev \dashv coev^R$.
