Answer: \(x>5a\) or \(x<-\dfrac{3a}{5}\).

Since both sides of the inequality are non-negative, we may square both sides:

\(|3x-a|>2|x+2a|\)

gives

\((3x-a)^2>4(x+2a)^2.\)

Expanding,

\(9x^2-6ax+a^2>4x^2+16ax+16a^2.\)

So

\(5x^2-22ax-15a^2>0.\)

Now factorise:

\(5x^2-22ax-15a^2=(5x+3a)(x-5a).\)

The critical values are

\(x=-\frac{3a}{5}\quad\text{and}\quad x=5a.\)

Since the quadratic has positive leading coefficient, it is positive outside the two roots. Therefore

\(x<-\frac{3a}{5}\quad\text{or}\quad x>5a.\)