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(Note: this has been rewritten to reflect the comments below).

The answer to #1 is basically yes, because the proof that the Lindenbaum algebra above T is atomless is completely constructive.

Start with T, and assume that T + ~Con(T) is consistent. Then there is a sentence S such that T + ~Con(T) neither proves nor disproves S (using the first incompleteness theorem via Rosser's trick). So T + ~Con(T)$\land$S is stronger than T + ~Con(T), but is still consistent. This means that T + ~(Con(T)$\lor$~S) is consistent, so T + Con(T)$\lor$~S is stonger than T.

If T $\vdash$ (Con(T)$\lor$S) $\to$ Con(T) then T $\vdash$ S $\to$ Con(T). But this means T $\vdash$ ~Con(T) $\to$ ~S which is impossible. This shows that T + (Con(T)$\lor$S) < T+ Con(T) .

So we can let T^H be T + (Con(T)$\lor$S).

(Note: this has been rewritten to reflect the comments below).

The answer to #1 is basically yes, because the proof that the Lindenbaum algebra above T is atomless is completely constructive.

Start with a (consistent) theory T to which the second incompleteness theorem applies, which means that T + ~Con(T) is also consistent. Then there is a sentence S such that T + ~Con(T) neither proves nor disproves S (using the first incompleteness theorem via Rosser's trick). So T + ~Con(T)$\land$~S is stronger than T + ~Con(T), but is still consistent. This means that T + ~(Con(T)$\lor$S) is consistent, so T + Con(T)$\lor$S is stonger than T.

If T $\vdash$ (Con(T)$\lor$S) $\to$ Con(T) then T $\vdash$ S $\to$ Con(T). But this means T $\vdash$ ~Con(T) $\to$ ~S which is impossible. This shows that T + (Con(T)$\lor$S) < T+ Con(T) .

So we can let T^H be T + (Con(T)$\lor$S).

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 RCA₀ and RCA₀ + "There is at least one noncomputable set".

Note:

If T + Cont(T) is inconsistent, the theory $T^H$ here is still well defined but is equivalent to $T$. The construction in this case just bounces back and forth between $T$ itself and the maximal element of the algebra (the one corresponding to an inconsistent theory).

One way to resolve this is to read the original question as only being about true theories, which is how I originally read it. If a true theory is consistent, then it is certainly consistent with Con(T). Moreover, the theory $T^H$ constructed here is a true theory if $T$ is.

(Note: this has been rewritten to reflect the comments below).

The answer to #1 is basically yes, because the proof that the Lindenbaum algebra above T is atomless is completely constructive.

Start with T, and assume that T + ~Con(T) is consistent. Then there is a sentence S such that T + ~Con(T) neither proves nor disproves S (using the first incompleteness theorem via Rosser's trick). So T + ~Con(T)$\land$S is stronger than T + ~Con(T), but is still consistent. This means that T + ~(Con(T)$\lor$~S) is consistent, so T + Con(T)$\lor$~S is stonger than T.

If T $\vdash$ (Con(T)$\lor$S) $\to$ Con(T) then T $\vdash$ S $\to$ Con(T). But this means T $\vdash$ ~Con(T) $\to$ ~S which is impossible. This shows that T + (Con(T)$\lor$S) < T+ Con(T) .

So we can let T^H be T + (Con(T)$\lor$S).

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 RCA₀ and RCA₀ + "There is at least one noncomputable set".

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 RCA₀ and RCA₀ + "There is at least one noncomputable set".

Note:

If T + Cont(T) is inconsistent, the theory $T^H$ here is still well defined but is equivalent to $T$. The construction in this case just bounces back and forth between $T$ itself and the maximal element of the algebra (the one corresponding to an inconsistent theory).

One way to resolve this is to read the original question as only being about true theories, which is how I originally read it. If a true theory is consistent, then it is certainly consistent with Con(T). Moreover, the theory $T^H$ constructed here is a true theory if $T$ is.