A ring has ITI with respect to an ideal $\mathfrak{a}$ if and only if it has ITR with respect to $\mathfrak{a}$.
A ring does not necessarily have ITI with respect to an ideal of finite type. (Note that the ideal in Quý's example is not of finite type.)
A ring that has ITI with respect to every ideal of finite type does not necessarily have ITI with respect to every ideal.
Proofs of the above, concrete examples, and further details on the ITI-property will be found in a joint work with P.H.Quý (who gave the accepted answer), available in due time. See also this question.