First of all, if you are doing guesswork plugging in numbers for x, then always always consider fractions less than one as one possibility. Often those numbers behave differently from other numbers, and can be a conclusion-changer in DS.
As for this particular question, I'll do an algebraic solution in case you are interested.
Question: is x > 0
Statement #1: |x+3|<4
That means that -4 < x + 3 < 4
Subtract 3 from all three conditions, and we have: -7 < x < 1. So, under this condition, x could be negative (down to -7) or positive (between zero and 1). Remember, there's a whole infinity of numbers between zero and 1: please don't forget they're there. By itself, Statement #1 is insufficient.
Statement #2: |x-3|<4
That means that -4 < x - 3 < 4
Add three to all three conditions, and we have -1 < x < 7. So, under this condition, x can be positive (up to 7) or negative (between -1 and 0). Again, another whole infinity of numbers between -1 and zero: don't forget about them! By itself, Statement #2 is insufficient.
Combined Statements #1 & #2
We combine the conditions -7 < x < 1 and -1 < x < 7. To satisfy both of those conditions simultaneously, x must be between -1 and +1: -1 < x < 1. That's the overlap, or intersection, of the two regions. Well, x can still be positive or negative, so even combined, the statements are insufficient. Answer = E.
Does that make sense? Please let me know if you have any questions on what I've said.
Mike
