I believe the first one should be written as 7.71x10^2 since there are 3 sig figs in 5.25x10^2, and the second one is right it has 3 sig figs. It is written as 3.40x10^2 because it has been rounded up from 3.39. You are allowed to round up your final answer to one place and only your final answer.
But what the hell do I know I've lost so many marks last year from sig figs...bummer.
If you were multiplying or dividing, Mozart would be right. Addition and subtraction with significant figures are tricky things, though.
The thing you have to keep in mind is that significant digits tell you about how accurately you think you have determined a number. For instance, suppose you have a scale accurate to the nearest tenth of a gram. You're working with something which happens to mass exactly 60.03 grams. Your scale might very well show it to you as 60.0 grams or 59.9 grams or 60.1 grams. In this case, your least significant digit is the tenths - that's the one about which you're most uncertain.
Suppose you then added .0002 g of something to the scale. Would the scale change its reading? Probably not. Does the fact that you've suddenly put something with only one significant digit on the scale change your three s.d. figure of 60.0 to only one s.d.? Also, no. The thing is that the three zeros in .0002 are not significant - they're just place holders.
Consider your first problem: 8.231 x 10^3 = 8231, four s.d., accurate to the one's place. 5.25 x 10^2 = 525, accurate to the one's place.. Now: you know the first number to the one's place, meaning you have no idea what's in the tenth's place or lower on that one. You also know the second one to the one's place. When you subtract them, you're subtracting as (and forgive the formatting):
8231.xxxxx
0525.xxxxx
where the x's indicate digits you have no information on. Our answer, then, will be accurate to the one's place, because we know what those digits "are" in both numbers. Therefore, 7706 accurate to the one's place, or 7.706 x 10^3.
In the second case, 350 is accurate only to the tens place. 10.5 is accurate to the tenths place. You know the second number far more accurately than the first. So, using the same convention:
35x.xxx
010.5xx
I can subtract the 0 from the three because both are significant. I can subtract the 1 from the 5 for the same reason. But I can't subtract the one's places or the tenth's places because I only know one of the digits in each place. What do I get when I subtract five from 'something'? It could be anything, so it's not significant. As for the hundredth's place and on down, I don't know either digit, so I certainly can't subtract them.
Therefore, 350 - 10.5 = 340, accurate to the tens place.
Think of it like this - your answer can only be as accurate as the least accurate number you use. In addition and subtraction, it's the place of the digits that matters. Subtracting a number accurate to the tenth's place from one accurate to the one's place gives an answer accurate only to the one's place, no matter how many significant digits are involved.