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The part (a) is not answered here - so let me include it, for the sake of completeness.$\newcommand{\inv}[1]{{#1}^{-1}}$
We want to show that $\inv{(g\circ f)}(Z)=\inv f(\inv g(Z))$. For any $w\in W$ we have:
$w\in \inv{(g\circ f)}(Z)$ $\Leftrightarrow$ $(g\circ f)(w)\in Z$ $\Leftrightarrow$ $g(...