Both sides are nonnegative, so squaring preserves the strict inequality: \[(3x-a)^2>4(x+2a)^2.\] Subtract and factor as a difference of squares: \[(3x-a-2x-4a)(3x-a+2x+4a)>0,\] \[(x-5a)(5x+3a)>0.\] As a>0, the critical values satisfy \(-3a/5<5a\). The product is positive outside these roots. Thus \[\boxed{x<-\frac{3a}{5}\quad\text{or}\quad x>5a}.\] The endpoints are excluded.