jmc

Setting the expressions for $g(x)$ given in the first two equations equal to each other gives us $3x+2=2f^{-1}(x)$, so $f^{-1}(x)=\frac{3x+2}{2}$. Substituting $f(x)$ into our expression for $f^{-1}$, we get $$\begin{array}{rl}\frac{3f(x)+2}{2}& =f^{-1}(f(x))\\ \Rightarrow \frac{3f(x)+2}{2}& =x\\ \Rightarrow 3f(x)& =2x-2\\ \Rightarrow f(x)& =\frac{2x-2}{3}.\end{array}$$Thus, $a=\frac{2}{3}$ and $b=\frac{-2}{3}$, so $\frac{a+b}{2}=0/2=\boxed{0}$.