Write and factorize the denominator.
note that 3x^2 - 3x - 5 = 3(x^2 - 2x - 3) + 3x + 4.
so (3x^2 - 3x - 5)/(x^2 - 2x - 3) = 3 + (3x + 4)/(x^2 - 2x - 3).
clearly, this will be > 3 iff (3x + 4)/(x^2 - 2x - 3) > 0.
now (3x + 4)/(x^2 - 2x - 3) = (3x + 4)/[(x - 3)(x + 1)].
for this to be positive, the numerator and the denominator must both have the same sign.
suppose 3x + 4 > 0, that is x > -4/3.
if -4/3 < x < -1, then x+1 is negative, and x-3 is also negative, so the denominator and numerator are both positive.
if -1 < x < 3, then x+1 is positive, and x-3 is negative, so the denominator is negative, so these values won't work.
if 3 < x, then x+1 is positive, and x-3 is positive, so the denominator and the numerator are both positive.
now, suppose x < -4/3. then x+1 and x-3 are always negative, so the denominator is positive, so these values won't work.
so the only values of x for which (3x^2 - 3x - 5)/(x^2 - 2x - 3) > 3 are in the intervals (-4/3,-1), and (3,∞).