5&6

5
Since you scale before you shift, scale the x-values by 1/2, so you have
sqr rt (2x) Move it to the left 6, sqr rt (2(x+6))=2x+12

6
Graph the equation, and calculate the minimum, using 0 as the lower bound. This happens at one hour, with a 1/3 percent oxygen. This is the lowest possible amount of oxygen in the stream.
(O(4)-O(2))/2 gives you the average rate of change over the interval [2,4]
6.52685 is the answer. This is how much, on average, the percent oxygen in the stream increases in an hour in between the second and fourth hour.