The first one is an awesome problem as you need many observations in order to solve it. I am going to describe the solution very briefly.

Let us fix an optimal solution (that is a choice of b[i] ≥ a[i] such that their xor is 0 and is minimized). Then we make the following observations:

1. We can assume that if a[i] ≤ a[j] then b[i] ≤ b[j].
2. For any i let us define x[i] as the largest bit in a[i]^b[i] (define it as  - 1 if a[i] = b[i]). Each positive value of x[i], apart from the largest one, is assumed at most one time. The largest one can be assumed at most twice.
3. For any i it holds x[i] ≤ 31.

With this observations (that I leave to prove to the reader) it is easy to create a solution. Indeed, once we have fixed the largest value of x[i], the previous observations are enough to uniquely define an optimal strategy (fixing greedily all bits from the largest to the smallest).

In some sense the described algo is almost greedy, indeed, after a single choice, a greedy algorithm works.

I think the second problem is this problem: http://mirror.codeforces.com/contest/429/problem/E. I have a comment about it here: http://mirror.codeforces.com/blog/entry/12265?#comment-169310