= Solution
\b[For the pure orders, yes.] Both structures satisfy the theory of <dense linear orders without endpoints>. By <quantifier elimination for dense linear orders without endpoints>, each formula is equivalent to a quantifier-free order formula. Inclusion of the rationals into the reals preserves and reflects all such formulas, hence is elementary:
$$
\boxed{(\mathbb Q,<)\preccurlyeq(\mathbb R,<).}
$$
\b[With a common named unit, yes.] In general an <elementary substructure> remains elementary when elements already in the <substructure> are named by new constants. A formula using the new constant is just an old-language formula with that common element as a parameter. Since $1\in\mathbb Q$, this applies directly.
\b[With the real constant interpreted by $\pi$, no choice inside the rationals works for inclusion.] A <substructure> embedding must preserve every constant. The inclusion would send a rational interpretation $q$ to that same real $q$, whereas the target constant is $\pi\notin\mathbb Q$. Thus the expanded rational structure is not even a <substructure> of the expanded real one.
This last failure concerns the specified inclusion. There is an abstract <elementary embedding> carrying a chosen rational $q$ to $\pi$: the increasing map $r\mapsto r+\pi-q$ has that property, and its image is a dense order without endpoints inside the reals. <Quantifier elimination> makes the embedding elementary, but its image is a translated copy of the rationals rather than the original rational subset.
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