Remember that to get y=f'(x) from y=f(x) you differentiate the function. In the same way, you integrate y=f'(x) to get y=f(x). What this essentially means is you dont actually care what the graph y=f'(x) actually is, you care a hell of a lot more about the area under the graph, which for the most part in the question is given to you or is easily calculated (for 4<=x<=6).
So with b)ii), you can see that the definite integral of f'(x) has a maximum value of 4, remembering that area above the x-axis is positive and that area under the x-axis is negative. So basically you're looking for when the area on top of the graph is greatest, and its value when it is the greatest. Fortunately for you, it's given in the question
A similar thing happens with part iii), the rectangle next to A2 can be calculated to be 6 units squared (3x2), and since A1 and A2 cancel, you get -6.
Because you know that at x=6, f(x)=-6 (from part iii) and that f(x) has a maximum at x=2, and f(x)=4, and that f(x)=0 at x=4 (from the total signed area of f'(x) between 0 and 4), you can accurately draw the graph given in part iv, which is a parabola
I guess general tip here is to notice that when there's no equation, theres usually some other way to solve the question that's right in front of you that most times you won't even notice. Sometimes it doesnt have to be as easy as bringing out your integrals and your dxs. When you're given the derivative graph, I guess look for the area under the graph, and not at the graph itself, because more likely than not you're gonna find f(x) anyway!
hope this helps