First things first, we need to create a common denominator. Let's factor everything first and see if anything divides out

#(2(2x-1))/(3(x^2-2x-8)) + (3 xx 3)/(3(x^2-5x+4))#

#(2(2x-1))/(color(green)(3)color(green)((x-4))(x+2)) + (3 xx 3)/(color(green)(3)(x-1)color(green)((x-4)))#

Nothing divides out, but it looks like the denominators are mostly the same, except that the right side doesn't have #(x+2)#, and the left side is missing #(x-1)#. Let's multiply both sides by the term they're missing (remember, multiply both the numerator and denominator)

#(x-1)/(x-1) xx (4x+2)/(3(x-4)(x+2)) + (9)/(3(x-1)(x-4)(x+2)) xx (x+2)/(x+2)#

#(4x^2-4x+2x-2)/(3(x-4)(x+2)(x-1)) + (9x+18)/(3(x-1)(x-4)(x+2))#

Now that the denominators match, let's combine the numerators:

#(4x^2-2x-2+9x+18)/(3(x-4)(x+2)(x-1))#

That denominator could use some work...

#3(x-4)(x+2)(x-1)#

#(3x-12)(x+2)(x-1)#

#(3x^2+6x-12x-24)(x-1)#

#(3x^2-6x-24)(x-1)#

#3x^3-3x^2-6x^2+6x-24x+24#

#color(green)(3x^3-9x^2-18x+24)#

Now we have

#(4x^2+7x+16)/(color(green)(3x^3-9x^2-18x+24))#

I hope that helps!