Heya! Okidoke, first off, thanks for providing me with this paper! I should check out the AQA website more often, so that's good to have.
Second off, I'll help with the part a)i). I think it's a BS question. You're 100% right: it's total bull imo, and I would have lost the marks to be fair. MS is implying that they want the voltage and current reading from right before there is a dip, at about 0.55-0.57 volts. The current at this point is about 0.9, and that gives us the answer they want. My best explanation (not great) for this is: since the X*Y is equal to power for a given point, the power keeps increasing as we get a higher voltage; once we reach a certain point and the gradient decreases, voltage may keep increasing but current decreases, and the decrease in current, even with an increasing voltage, yields a smaller power. The increasing voltage can't compensate for the currents decrease, as the current will decrease at a greater factor than the PD. That's barely visible in the graph, but so be it.
For part a)iii), we calculated the power. The answer we may have gotten may not have been correct, but if they don't award ECF marks, then they should go drink a gallon of peach flavoured detergent. I'll explain this in fragments below, I feel as though you hate my wall of text.
Efficiency = useful/total, and we know that for an area of 1 metre, the total is 450W, assuming 100% efficiency. It's not 100%, it's 15%, 0.15. Thus, for an area of 1 metre, we get 15% of 450, about 67.5 if my head is fine with me. Divide your value for power by that, and you get the ratio of your cells area to the area of a metre at 15% efficiency.
How do you use that ratio? You tell me :P
Again, part a)i) was BS imo. Others can argue it isn't and that the average student will have the logic to note that the current decreases far faster than the increase in p.d can compensate at the point where the current decreases, but still, COME ON