The answer to #1 is basically yes, because the proof that the Lindenbaum algebra above T is atomless is completely constructive.  This assumes you replace Q with I-$\Sigma^0_1$ per my comment above.  By luck I was just discussing this a few days ago so I have it in my head. 

Start with T.  Since Con(T) is not an atom, ~Con(T) is not a coatom, and in particular the second incompleteness theorem shows that (T + ~Con(T)) + Con(T + ~Con(T)) is stronger than T + ~Con(T). So we just take the complement of this in the Lindenbaum algebra: let $T^H$ be T + [Con(T)$\lor$~Con(T + ~Con(T))]. 

One cute application of this technique is to construct a concrete system strictly between RCA<sub>0</sub> and RCA<sub>0</sub> + "There is at least one noncomputable set".