Let 9n/10 be sorted and n/10 be not sorted part.
For insertion sort:
1st 9n/10 elements will take 1 comparison each(since sorted) = 9n/10 comparison.
Starting from 9n/10+1 element it’ll in worst case take 9n/10 comparison(smaller than all)
9n/10+2 element will take 9n/10+1
Total comparison = 9n/10 (first 9n/10) + (9n/10) + (9n/10+1) + (9n/10 + 2) + …… (9n/10 + n/10-1)
If you sum it you’ll get O(n^2) which is less than merge sort which has O(nlogn).
P.S.: Are my answers correct, just for checking.