Answer
$ a.\quad f^{-1}(x)=5x-35$
$ b.\quad$ see image
$ c.\quad \displaystyle \left.\frac{df}{dx}\right|_{x=-1}=\frac{1}{5},\qquad \left.\frac{df^{-1}}{dx}\right|_{x=34/5}=5$
Work Step by Step
$ a.\quad$
Set $y=f(x)$. Interchange x and y
$y=\displaystyle \frac{1}{5}x+7$
$x=\displaystyle \frac{1}{5}y+7$
... solve for y
$x-7=\displaystyle \frac{1}{5}y$
$y=5x-35$
... replace y with $f^{-1}(x)$
$f^{-1}(x)=5x-35$
$ b.\quad$
see graphs (attached image)
$ c.\quad$
$f(x)=\displaystyle \frac{1}{5}x+7$
$\displaystyle \frac{df}{dx}=\frac{1}{5},\qquad \left.\frac{df}{dx}\right|_{x=-1}=\frac{1}{5}$
$f(-1)=\displaystyle \frac{1}{5}(-1)+7=\frac{34}{5}\quad\Rightarrow\quad f^{-1}(\frac{34}{5})=-1$
$f^{-1}(x)=5x-35$
$\displaystyle \frac{df^{-1}}{dx}=5,\qquad \left.\frac{df^{-1}}{dx}\right|_{x=34/5}=5$
