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)=\dfrac{3x+2}{2} $. Substituting $ f(x) $ into our expression for $ f^{-1} $, we get \begin{align*} \dfrac{3f(x)+2}{2}&=f^{-1}(f(x)) \\ \Rightarrow \dfrac{3f(x)+2}{2}&=x \\ \Rightarrow \quad 3f(x)&=2x-2 \\ \Rightarrow \quad f(x)&=\frac{2x-2}{3}\end{align*} Thus, $ a=\frac{2}{3} $ and $ b=\frac{-2}{3} $, so $ \dfrac{a+b}{2}=0/2=\boxed{0} $.