 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 $i\in \{1,\dots ,5\}$, take the identity and the three double transpositions fixing $i$. For instance, the subgroup fixing $5$ isThese five point stabilizers are precisely the Sylow $2$-subgroups. Indeed, the $15$ double transpositions occur three to each such subgroup, so there areof them. Together with the previous parts, this gives the Sylow subgroups of S3, S4 and A5.