$$\begin{aligned} & f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}=y, 2 x^2+3 x+8>0 \quad \forall x \in \mathbb{R} \\ & \Rightarrow \quad x^2(2 y-2)+x(3 y+3)+8 y-8=0 \end{aligned}$$

Since $$x \in \mathbb{R}$$, the equation has real roots

$$\begin{aligned} & \Rightarrow \quad D \geq 0 \\ & \Rightarrow(3 y+3)^2-4(2 y-2)(8 y-8) \geq 0 \\ & \Rightarrow 9(y+1)^2-64 y(y-1)^2 \geq 0 \\ & \Rightarrow(3 y+3)^2-(8 y-8)^2 \geq 0 \\ & \Rightarrow(11 y-5)(-5 y+11) \geq 0 \\ & \Rightarrow\left(y-\frac{5}{11}\right)\left(y-\frac{11}{5}\right) \leq 0 \\ & \Rightarrow y \in\left[\frac{5}{11}, \frac{11}{5}\right] \end{aligned}$$

Sum of maximum and minimum value

$$\begin{aligned} & y_{\max }+y_{\min }=\frac{5}{11}+\frac{11}{5}=\frac{25+121}{55} \\ & =\frac{146}{55}=\frac{m}{n} \Rightarrow m+n=201 \end{aligned}$$