jmc

Since $f$ is the function that adds one, $f^{-1}$ is the function that subtracts one. Since $g$ is the function that doubles, $g^{-1}$ is the function that halves. This lets us compute from inside out:

$$\begin{array}{rlr}& f(g^{-1}(f^{-1}(f^{-1}(g(f(5))))))& \\ & =f(g^{-1}(f^{-1}(f^{-1}(g(6)))))& \text{added 1}\\ & =f(g^{-1}(f^{-1}(f^{-1}(12))))& \text{doubled}\\ & =f(g^{-1}(f^{-1}(11)))& \text{subtracted 1}\\ & =f(g^{-1}(10))& \text{subtracted 1}\\ & =f(5)& \text{half}\\ & =\boxed{6}& \text{added 1}.\end{array}$$