The answer to the second question is negative. Consider a (first-order) theory of PA without the successor axiom like fpa. Then T is consistent and proves its own consistency. Let k be the complexity of this proof of consistency. Then trivially T is k-consistent and k-proves its k-consistency.

The answer to the second question is negative. Consider a (first-order) theory of PA without the successor axiom like fpa. Then T is consistent and proves its own consistency. Let k be the complexity of this proof of consistency. Then trivially T is k-consistent and k-proves its k-consistency.

[the following is added in an edit] As my comment indicated, the previous remarks are not completely correct. Instead: Let T be a (first-order) theory of PA without the successor axiom like fpa. Then T is consistent and proves its own consistency. In fact, T proves:

(x)(there does not exist an x-proof of a contradiction in T). (*)

Let this proof have complexity z. Then T can prove - call this sentence S(y) -

"there does not exist a y-proof of a contradiction in T" for any y,

but its proof may have greater complexity than z; indeed, if the proof uses (*), then it will have greater complexity, by say f(y). That is, T proves S(y) with a proof of complexity no greater than z + f(y). Suppose there exists a k such that k <= z + f(k). Then T k-proves S(k), i.e. T k-proves the k-consistency of T. Since T is k-consistent, the answer to the question would therefore be negative.

Is there likely to exist k <= z + f(k)? The important step is to find k which can be referred to with complexity much less than k. If complexity is defined to be say the length of the proof, then this should be possible by defining an exponential operator and defining k to be a power of two reasonably large numbers.