= Solution
A Sylow $2$-subgroup of $A_5$ has order four. Since $A_5$ has no element of order four, each is a Klein four-group. For every $i\in\{1,\ldots,5\}$, take the identity and the three double transpositions fixing $i$. For instance, the <subgroup> fixing $5$ is
$$
\{1,(12)(34),(13)(24),(14)(23)\}.
$$
These five point stabilizers are precisely the Sylow $2$-subgroups. Indeed, the $15$ double transpositions occur three to each such <subgroup>, so there are
$$
\boxed{n_2=5}
$$
of them. Together with the previous parts, this gives the <Sylow subgroups of S3, S4 and A5>.
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