= Solution
Apply the <Friedman–Moschovakis coding lemma> with
$$
\lambda=\aleph_2.
$$
The given map $\pi:\omega^\omega\twoheadrightarrow\aleph_2$ supplies the required real codes for ordinals below $\lambda$, while each $g_\xi:\omega^\omega\twoheadrightarrow\mathcal P(\xi)$ supplies codes for all possible initial segments of a subset of $\lambda$.
For completeness, fix $A\subseteq\aleph_2$ and form the associated <Friedman–Moschovakis coding game>. The players use $\pi$ to announce ordinals and $g_\xi$ to announce candidate codes for $A\cap\xi$, while each may challenge the other's code at a larger ordinal. The Friedman–Moschovakis diagonal argument shows that Player I cannot have a winning strategy. By the <axiom of determinacy>, Player II has one. The coherence tests in the game ensure that a fixed winning strategy for II can belong to at most one set $A$: if it purported to code distinct $A$ and $B$, a play reaching an ordinal above the least point of disagreement would defeat it.
Every strategy for a game on $\omega$ is coded by a real. Define $F:\omega^\omega\to\mathcal P(\aleph_2)$ by sending a code for a winning II-strategy to the unique set that it determines, and sending all other reals to the empty set. Every $A\subseteq\aleph_2$ has such a strategy, so $F$ is surjective. Hence
$$
\boxed{\omega^\omega\twoheadrightarrow\mathcal P(\aleph_2).}
$$
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