To 1 if it's the same algorithm running on the same machine then you could probably assume that it just needs to be cut in half.
2) I'm not going to directly answer this but I will say that
O(N^2) is just a notation to say that all 'certain kinds' of functions belong to the set O(N^2) which means that all of these functions are growing at the same rate towards their upper bound because O is the upper bound.
So basically you could say that an algorithm who's time analysis is 0.015n^2+2n+4 belongs to the set O(n^2) and at the same time an algorithm who's time analysis is 1000000n^2+90n+15 also belongs to the set O(n^2). It means both of these algorithms belong to O(n^2).
For 1, I don't think it wouldn't take half the time, it would take a quarter, its O(N^2) not O(N2).
n = 100, s = 40;
n * n = 10,000
10,000 / s = 250 n per second
n = 50, s = ?
50 * 50 = 2,500
s = 2,500 / 250 n per second = 10;
so it would take 10 seconds.
as for converting binary to decimal, its really simple, binary starts from the right and moves left, if its 0 you ignore it, if its 1 then the 'value' of that 1 is 2^(x - 1), where x is the distance from 1 to the right-end of the binary number, all you have to do is add up all the values of the 1's.