$\frac{6 x^4+13 x^3+2 x^2-x+3}{2 x^2+3 x-2}=f(x)+\frac{A}{a x-1}+\frac{B}{x+b}$
$\begin{aligned} & =\frac{3\left(2 x^2+3 x-2\right) x^2+4 x^3+8 x^2-x+3}{2 x^2+3 x-2} \\ & =\frac{\left(3 x^2+2 x+1\right)\left(2 x^2+3 x-2\right)+5}{2 x^2+3 x-2} \\ & =3 x^2+2 x+1+\frac{5}{2 x^2+3 x-2} \\ & =\left(3 x^2+2 x+1\right)+\frac{5}{(2 x-1)(x+2)} \\ & =\left(3 x^2+2 x+1\right)+\frac{2}{2 x-1}+\frac{(-1)}{x+2} \\ & \therefore \quad f(x)=3 x^2+2 x+1 \\ & \quad A=2, a=2 \\ & \quad B=-1, b=2 . \\ & \therefore \quad f(1)+a \cdot B+b \cdot A=6+2(-1)+2(2)=8 .\end{aligned}$