1. No matter what, insertion sort ALWAYS does the same thing.  So, even if the data is basically sorted already, it still will do the same amount of work.

2. In a single pass of the data you can find the two elements that are out of place by looking for an element where:

some_list[i] > some_list[i+1]

Since it's just a single pass through the data, the runtime would be O(n), linear.