a heap with n elements that supports Insert and Extract-Min, Which of the following tasks can you achieve in O(logn) time?

It depends on what the running times are of the operations insert and extract-min. In traditional heaps, both take ϴ(log n) time. However, in finger-tree-based heaps, only insert takes ϴ(log n) time, while extract-min takes O(1) time. There, you can find the fifth smallest element in O(5) = O(1) time and the median in O(n/2) = O(n) time. You can also find the largest element in O(n) time.

Why is "Find the largest element stored in the heap."not correct, my understanding here is that you can use logN time to go to the bottom of the heap, and one of the element there must be the largest element.

The lowest level of the heap contains half of the elements. More correctly, half of the elements of the heap are leaves--have no children. The largest element in the heap is one of those. Finding the largest element of the heap, then, will require that you examine n/2 items. Except that the heap only supports insert and extract-min, so you end up having to call extract-min on every element. Finding the largest element will take O(n log n) time.

"Find the fifth-smallest element stored in the heap." this should take constant time right, because you only need to go down 5 layers at most?

This can be done in log(n) time. Actually 5*log(n) because you have to call extract-min five times. But we ignore constant factors. However it's not constant time because the complexity of extract-min depends on the size of the heap.

"Find the median of the elements stored in the heap." should this take O(n) time? because we extract min for the n elements to get a sorted array, and take o(1) to find the median of it?

The median is the middle element. So you only have to remove n/2 elements from the heap. But removing an item from the heap is a log(n) operation. So the complexity is O(n/2 log n) and since we ignore constant factors in algorithmic analysis, it's O(n log n).