for (ii), i think it will take O(n+klogn)........
since n/2 minimum elements are present in leaves of a max heap, now you have to find kth minimum among them, you can work it like this,
create min heap for that n/2 elements -> O(n)
perform extract_min for k times -> O(klogn)
in total it will take-> O(n+klogn), though if k is some constant, answer will be O(n), but in worst case k can be of O(n)..
therefore i think in worst case to find kth min in max heap will take -> O(n+klogn)