Let $\Sigma(u_1 \dots u_n), \Sigma'(u_1 \dots u_n)$ be complete $n$-types of an $\mathcal{L}$-theory $T$, where $\mathcal{L}$ is a countable language.
Q1: Is there a model $\mathcal{M} \models T$ such that $\mathcal{M}$ realizes both $\Sigma$ and $\Sigma'$?
Q2: If so, can we choose $\mathcal{M}$ to be countable?
I know a few theorems which seem like they might be relevant:
A principal complete $n$-type of $T$ must be realized by any model of $T$.
If $\Sigma$ is a complete $n$-type of $T$ and $\mathcal{M} \models T$, then there exists an elementary extension of $\mathcal{M}$ which realizes $\Sigma$.
If $\Sigma$ is a non-principal complete $n$-type of $T$ then there exists a countable $\mathcal{M} \models T$ such that $\Sigma$ is not realized in $\mathcal{M}$.
Assume $T$ is complete.
Let $M$ be an arbitrary model of $T$. Realize $\Sigma$ in an elementary extension $M\preceq M'$. Then realize $\Sigma'$ in an elementary extension $M'\preceq M''$. Done and done.
If you want the model to be countable, use Löwenheim-Skolem to find a countable elementary submodel of $M''$ containing the realizations of $\Sigma$ and $\Sigma'$.
If $T$ is not complete, then $\Sigma$ and $\Sigma'$ might disagree about $L$-sentences, in which case they obviously can't be simultaneously realized.