= Solution
By part ii every subset of the <Baire space of sequences> is $2^{\aleph_0}$-Suslin. If $2^{\aleph_0}\leq\aleph_2$, part i would make every such set $\aleph_2$-Suslin. Therefore
$$
\boxed{\Psi_{\aleph_2}\Longrightarrow 2^{\aleph_0}>\aleph_2.}
$$
Now suppose $2^{\aleph_0}>\aleph_2$. The <axiom of choice> gives a set $B\subseteq\omega^\omega$ of cardinality exactly $\aleph_2$. If $B$ were $\aleph_1$-Suslin, the <Aleph-one-Suslin decomposition into analytic sets> would write it as a union of $\aleph_1$ <analytic set>[analytic sets]. If all those analytic sets were countable, their union would have cardinality at most $\aleph_1$, so one of them is uncountable. The <perfect set property> for analytic sets then makes that member, and hence $B$, have cardinality $2^{\aleph_0}$, contradicting
$$
|B|=\aleph_2<2^{\aleph_0}.
$$
Thus $B$ is not $\aleph_1$-Suslin, and
$$
\boxed{2^{\aleph_0}>\aleph_2\Longrightarrow\Psi_{\aleph_1}.}
$$
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