= Solution
All six reflections square to the identity, and the rotation $R^3$ is the half-turn. These exhaust the nonidentity involutions: $R^k$ has square one only when $2k\equiv0\pmod6$, giving $k=3$ apart from the identity. Thus the order-two <subgroups> are
$$
\boxed{\{I,R^3\},\qquad\{I,m_j\}\quad(j=1,\ldots,6).}
$$
A reflection cannot have order three. The rotations of order three are $R^2$ and $R^4$, and they generate the same subgroup:
$$
\boxed{\{I,R^2,R^4\}=\langle R^2\rangle.}
$$
\b[All these subgroups are cyclic], because a group of prime order is cyclic; the displayed nonidentity elements supply their generators. This gives seven subgroups of order two and one of order three, consistently with <involutions in an even dihedral group> and the <classification of subgroups of a dihedral group>.
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