a) On [0,1], f(x) is non-negative so the integral can be at least 0. But we know that f(1/2) > 0. So because of continuity, there exists an e > 0 for which f(x) > 0 on the interval (1/2-e,1/2+e). Then:
Now, the first and third integrals may be 0, but the second one is > 0.
b) If f isn't continuous, then you can construct a function for which f(x) > 0 for x in (1/2-e,1/2+e) isn't true. For example; keep f(x) at 0 everywhere, except at x = 1/2.