\(a,\dfrac{3x+8}{x+2}+\dfrac{x}{x+2}=\dfrac{3x+8+x}{x+2}=\dfrac{4x+8}{x+2}=\dfrac{4\left(x+2\right)}{x+2}=4\)
\(b,MTC=2x-4\)
\(\dfrac{\left(x+3\right)^2}{x-2}-\dfrac{2x^2+2x+13}{2x-4}=\dfrac{\left(x+3\right)^2.2}{\left(x-2\right).2}-\dfrac{2x^2+2x+13}{2x-4}\\ =\dfrac{\left(x+3\right)^2.2-\left(2x^2+2x+13\right)}{2x-4}=\dfrac{2x^2+12x+18-2x^2-2x-13}{2x-4}=\dfrac{10x-5}{2x-4}\)