 Solution

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:

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.

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.