= Solution
Suppose $T$ is an <omega-categorical theory>. By the <Ryll-Nardzewski theorem>, every finite-arity <complete type> over the empty set is isolated. Let $p(x,a)$ be a <complete type> over a finite tuple in the unique <countable> model $M$, and take an <elementary extension> realizing it at $b$. The joint type of $(b,a)$ is isolated by a formula $\theta(x,y)$.
The <complete type> of $a$ contains $\exists x\,\theta(x,y)$, since this formula is true of $a$ in the extension. Hence it is true of $a$ in $M$, and some $b'\in M$ satisfies $\theta(b',a)$. Isolation of the joint type implies that $b'$ realizes $p$. Thus the <countable> model is an <omega-saturated model>.
\b[The converse fails.] The complete <successor theory> from question 4 has a <countable> model
$$
M=\mathcal N\ \sqcup\ \bigsqcup_{j\in\mathbb N}\mathbb Z_j
$$
with countably infinitely many two-sided successor components. It is omega-saturated. Over finitely many parameters, a complete one-variable type either fixes a finite successor/predecessor translate of a parameter or zero, or asks for a point in no component represented by those parameters and zero. The first kind is realized in the named component, and the second in an unused integer component. The quantifier-elimination description of <classification of models of the successor theory> shows there are no other cases.
Nevertheless this theory is not omega-categorical: $M$ and its standard model $\mathcal N$ are <countable> and not isomorphic. Thus existence of a <countable> saturated model is strictly weaker than <countable> categoricity.
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