[tex]A)\\f(x)=5x+2\\\\f(7):\ \text{Put x = 7 to the equation of the function:}\\\\f(7)=5(7)+2=35+2=37\\\\\boxed{f(7)=37}[/tex]
[tex]B)\\f(x)=5x+2\to y=5x+2\\\\\text{exchange x to y}\\\\x=5y+2\\\\\text{solve for y}\\\\5y+2=x\qquad\text{subtract 2 from both sides}\\\\5y=x-2\qquad\text{divide both sides by 5}\\\\y=\dfrac{1}{5}x-\dfrac{2}{5}\\\\\boxed{f^{-1}(x)=\dfrac{1}{5}x-\dfrac{2}{5}}[/tex]
[tex]C)\\f^{-1}(x)=\dfrac{1}{5}x-\dfrac{2}{5}\\\\f^{-1}(7):\ \text{Put x = 7 to the equation of the function}\ f^{-1}(x):\\\\f^{-1}(7)=\dfrac{1}{5}(7)-\dfrac{2}{5}=\dfrac{7}{5}-\dfrac{2}{5}=\dfrac{7-2}{5}=\dfrac{5}{5}=1\\\\\boxed{f^{-1}(7)=1}[/tex]
[tex]D)\\f(f^{-1}(7))\\\\f^{-1}(7)=1,\ \text{Therefore put x = 1 to the equation of the function}\ f(x):\\\\f(1)=5(1)+2=5+2=7\\\\\boxed{f(f^{-1}(7))=7}[/tex]