The answer to (i) is yes. Otherwise, for some countable ordinal $\alpha$ we would have $=_\alpha\implies=_{\omega_1}$, and this is false. One way to see that is to use Scott's isomorphism theorem (see below) and the fact that isomorphism on countable structures is not Borel. Of course this takes a bit of work; with some care, you can also whip up concrete examples (certain pairs of ordinals together with additional labelled points; I believe Rosenstein's book Linear orderings, which is sadly out-of-print, has details).

The answer to (ii) is also yes. This is Scott's isomorphism theorem: for every countable structure $\mathfrak{A}$ there is an $\mathcal{L}_{\omega_1,\omega}$-sentence (= countable rank) $\varphi$, called a Scott sentence, such that the countable models of $\varphi$ are exactly the structures isomorphic to $\mathfrak{A}$.

Neither appearance of countability can be dropped:

• Since $\mathcal{L}_{\omega_1,\omega}$ has downward Lowenheim-Skolem for individual sentences, no uncountable structure can be pinned down up to isomorphism by a single $\mathcal{L}_{\omega_1,\omega}$-sentence.

• Some Scott sentences have countable models! Indeed, consider the linear order $(\mathbb{Q};<)$. This is potentially isomorphic (= is $\mathcal{L}_{\infty,\omega}$-equivalent, or by Barwise becomes isomorphic in some forcing extension) to any dense linear order without endpoints, including for example $(\mathbb{R};<)$. So a fortiori any $\mathcal{L}_{\omega_1,\omega}$-sentence true in $(\mathbb{Q};<)$ also has uncountable models.

The answer to (i) is yes. Otherwise, for some countable ordinal $\alpha$ we would have $=_\alpha\implies=_{\omega_1}$, and this is false. One way to see that is to use Scott's isomorphism theorem (see below) and the fact that isomorphism on countable structures is not Borel. Of course this takes a bit of work; with some care, you can also whip up concrete examples (certain pairs of ordinals together with additional labelled points; I believe Rosenstein's book Linear orderings, which is sadly out-of-print, has details).

The answer to (ii) is also yes. This is Scott's isomorphism theorem: for every countable structure $\mathfrak{A}$ there is an $\mathcal{L}_{\omega_1,\omega}$-sentence (= countable rank) $\varphi$, called a Scott sentence, such that the countable models of $\varphi$ are exactly the structures isomorphic to $\mathfrak{A}$.

Neither appearance of countability can be dropped:

• Since $\mathcal{L}_{\omega_1,\omega}$ has downward Lowenheim-Skolem for individual sentences, no uncountable structure can be pinned down up to isomorphism by a single $\mathcal{L}_{\omega_1,\omega}$-sentence.

• Some Scott sentences have uncountable models! Indeed, consider the linear order $(\mathbb{Q};<)$. This is potentially isomorphic (= is $\mathcal{L}_{\infty,\omega}$-equivalent, or by Barwise becomes isomorphic in some forcing extension) to any dense linear order without endpoints, including for example $(\mathbb{R};<)$. So a fortiori any $\mathcal{L}_{\omega_1,\omega}$-sentence true in $(\mathbb{Q};<)$ also has uncountable models.