This is part of the HW set. The help you need has been given in the previous problem. You basically need to use the gaussian distribution. For part B the answer is very similar to part d of the the previous problem. Now x in the gaussian represents the excess of forward(or backward) steps over N/2. So in terms of x and l (l is the length of a step), the net distance traveled should be 2xl, (the excess of forward steps of N). In this case, the gaussian distribution is . The width is , that is steps of length l. Now you can answer question 2.25 b.
Question 2.25 c is difficult and I'm giving here the answer:
In page 42, the book shows that the mean free path l in air is ~ 150 nm, and the average collision time is s. In 1 second we have that N~ steps, and the net distance traveled is =80000 steps or 12 nm.
From here we can see that N increases with time linearly by the distance traveled goes as , so goes as . At higher T the mean free path, which is proportional to V/N (we saw this in class) would increase in proportion to T. But the velocity is also larger, in proportion to . so the collision time increases only in proportion to . The expected distance traveled is proportional to times the mean free path, which makes the expected distance traveled to increase as . This is the correct result.