Hint: Build an elementary chain $(M_\alpha)_{\alpha<\aleph_1}$ of countable models, with $M_\alpha$ a proper elementary substructure of $M_\beta$ for all $\alpha<\beta$, and such that each $M_\alpha$ is isomorphic to the prime model $M$. Show that the union of this chain is atomic and has cardinality $\aleph_1$.
To handle the limit step of the transfinite construction, you will need to use the fact that a countable model is atomic if and only if it is prime, and countable prime models are unique up to isomorphism if they exist.
For 1 you already noted that $T_1 \cup T_2$ must be inconsistent. Following Chris Eagle's hint from the comments we can apply compactness to find $\phi \in T_1$ and $\psi \in T_2$ such that $\{ \phi, \psi \}$ is inconsistent (this assumes that $T_1$ and $T_2$ are closed under conjunctions, but we may as well assume that). We claim that $\phi$ axiomatises $T_1$ and $\psi$ axiomatises $T_2$. Every model of $T_1$ is a model of $\phi$. Now let $M \models \phi$, and suppose that $M \not \models T_1$. Then by assumption $M \models T_2$, and so $M \models \psi$. This contradicts $\{ \phi, \psi \}$ being inconsistent, so we must have $M \models T_1$. We see that $\phi$ axiomatises $T_1$. Analogously we find that $\psi$ axiomatises $T_2$.
For 2 I assume that "closed" means "closed under consequence". We first prove that $T$ is closed. Let $\phi$ be a consequence of $T$. Then $\{ \neg \phi \} \cup T$ is inconsistent, so by compactness there are $\psi_1, \ldots, \psi_n \in T$ such that $\{\neg \phi, \psi_1, \ldots, \psi_n\}$ is inconsistent. There must then be some $k < \omega$ such that $\psi_1, \ldots, \psi_n \in T_k$. It follows that $\phi$ is a consequence of $T_k$ and so $\phi \in T_k$ as $T_k$ is closed. We conclude that $\phi \in T$, as required.
Finally, to see that $T$ is not finitely axiomatisable we use TomKern's hint from the comments. Suppose for a contradiction that $T$ is finitely axiomatisable. Let $\{\phi_1, \ldots, \phi_n\}$ be a finite axiomatisation of $T$. Write $\phi$ for $\phi_1 \wedge \ldots \wedge \phi_n$, so $\phi$ axiomatises $T$ and hence $\phi$ is a consequence of $T$. We saw before that $T$ is closed, so $\phi \in T$. That means that $\phi \in T_k$ for some $k < \omega$. We then get $T_k = T$, because $\phi$ implies every sentence in $T$ and $T_k$ is closed. We thus have that $T = T_k \subsetneq T_{k+1} \subseteq T$, which is a contradiction. So $T$ cannot be finitely axiomatisable.
Best Answer
I don't want to give away the whole problem, but here are a couple hints to get started:
First, show that if $\Sigma$ is a finite language then for each $n$ there are only finitely many $\Sigma$-structures of cardinality $\le n$ (up to isomorphism). In fact, more is true: every finite $\Sigma$-structure is characterized up to isomorphism by a single $\Sigma$-sentence. This isn't needed here, though.
Next, remember that by the compactness theorem, if a theory has arbitrarily large finite models then it has infinite models.
So suppose $(T_i)_{i\in\mathbb{N}}$ is an increasing sequence of closed theories such that $T:=\bigcup T_i$ has no infinite model. Then some $T_i$ satisfies "there are at most $n$ elements in the domain" for some $n$; in particular, by the first bulletpoint there is some finite set of structures $\{\mathcal{M}_1,...,\mathcal{M}_k\}$ such that every model of $T_i$ is isomorphic to one of the $\mathcal{M}_j$s.
Now, suppose $S_2$ is a closed theory properly containing $S_1$. What can you say about the set of models of $S_2$ versus the set of models of $S_1$? Looking back to the previous bulletpoint, what does this say about any increasing sequence $(T_i)_{i\in\mathbb{N}}$ of closed theories whose union has no infinite models?