$$\begin{aligned} & f(x)+3 f\left(\frac{24}{x}\right)=4 x, x \neq 0 \quad \ldots(1)\\ & \text { replace } x \text { by } \frac{24}{x} \\ & f\left(\frac{24}{x}\right)+3 f\left(\frac{24}{24}\right)=4\left(\frac{24}{x}\right)=\frac{96}{x} \quad \ldots(2) \\ & 3 \times(2)-(1) \\ & \Rightarrow 8 f(x)=\frac{96.3}{x}-4 x \Rightarrow f(x)=\frac{36}{x}-\frac{x}{2} \\ & f(3)+f(8)=\left(12-\frac{3}{2}\right)+\left(\frac{36}{8}-4\right) \end{aligned}$$

$=8+\frac{36}{8}-\frac{12}{8}=11$