jmc

Since $f$ is the function that adds two, $f^{-1}$ is the function that subtracts two. Since $g$ is the function that divides by $3,$ $g^{-1}$ is the function that triples. This lets us compute from inside out: $$\begin{array}{rll}& f(g^{-1}(f^{-1}(f^{-1}(g(f(19))))))& \\ & =f(g^{-1}(f^{-1}(f^{-1}(g(21)))))& \text{added 2}\\ & =f(g^{-1}(f^{-1}(f^{-1}(7))))& \text{divided by 3}\\ & =f(g^{-1}(f^{-1}(5)))& \text{subtracted 2}\\ & =f(g^{-1}(3))& \text{subtracted 2}\\ & =f(9)& \text{tripled}\\ & =\boxed{11}& \text{added 2}\end{array}$$