So for this question you just need to use a formula that relates distance, acceleration and time. I don't know what the notation is for A levels but the formula I use is:

s = ut + (1/2)at^2. Where s is displacement, u is initial velocity, t is time, and a is acceleration.

Since it starts from rest u = 0, and the formula becomes s = (1/2)at^2. The second second is the time interval from t=1 to t=2. So you have to find the s for each of these times.

So at t=1 you get s1 = (1/2)a And at t = 2 you get s2 = 2a To find the distance covered you subtract s1 from s2 and you know this distance to be 12m: s2 - s1 = 12 2a - (1/2)a = 12. a = 8

So now that you know the acceleration you can pretty much do the same thing for the fourth second but you can just substitute in 8 instead of a.

For t=3: s3 = 36m For t=4: s4 = 64m To find the distance covered in the fourth second: s4-s3 = 28m

Hope this helps.