Fields the closure of which is $\mathbb{C}$

Yes to the first question, no to the other one. This answer needs some theory of trascendental extensions, though. $\Bbb R$ is algebraic over some purely trascendental extension of $\Bbb Q$ by $\beth_1$-many variables: namely, there is a field $\Bbb Q\subseteq k\subseteq \Bbb R$ such that $\overline k\supseteq \Bbb R$ and such that $k\cong \Bbb Q(X_\alpha\,:\,\alpha\in\beth_1)$, the field of rational functions in uncountably many variables.

Now, $\Bbb R$ cannot be $k$, because there are several ring homomorphisms $f:k\to k$ such that $\left. f\right\rvert_{\Bbb Q}=id$. On the other hand, any homomorphism of rings $\Bbb R\to\Bbb R$ which fixes $\Bbb Q$ must be the identity.

The least (or even a minimal) field $k$ such that $\overline k=\Bbb C$ does not exist: as before, such a field $k$ ought to contain $k'=\Bbb Q(\xi_\alpha\,:\,\alpha\in\beth_1)$ with $\overline{k'}=\Bbb C$ and $\xi_i\in k$ algebraically independent over $\Bbb Q$. By minimality, $k=k'$. Now, $k''=\Bbb Q(\xi^2_\alpha\,:\,\alpha\in\beth_1)$ is strictly contained in $k'$ and it has the same algebraic closure. At a closer inspection, the only fields which are minimal in the family of fields which have their same algebraic closure are $\Bbb Q$ and $\Bbb F_p$ (where $p$ is a prime).

According to the Axiom of Choice, there is a field automorphism $\tau$ of $\mathbb C$ that does not map $\mathbb R$ into itself. Then $F := \tau(\mathbb R)$ has the property that $\overline{F} = \mathbb C$.

It is interesting that such a field $F$ cannot be a Borel set in $\mathbb C$, however. It cannot be constructed (and proved correct) in ZF set theory only.