Answer
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Work Step by Step
$\underline{\text{To show}\ \small E\left(\dfrac{W-\mu}{\sigma} = 0\right):}$
Write $\small \dfrac{W-\mu}{\sigma} = \dfrac{1}{\sigma}W + \left(-\dfrac{\mu}{\sigma}\right)$ then use Corollary 3.5.1 (p. 148) with $\small a=\dfrac{1}{\sigma}$ and $\small b=-\dfrac{\mu}{\sigma}$ to get
$\begin{align*}
E\left(\dfrac{W-\mu}{\sigma}\right) &= E\left(\frac{1}{\sigma}W + \left(-\frac{\mu}{\sigma}\right)\right) \\
&= \frac{1}{\sigma}E(W) + \left(-\frac{\mu}{\sigma}\right) \quad [\ \text{Corollary 3.5.1}\ ] \\
&= \frac{1}{\sigma}\mu + \left(-\frac{\mu}{\sigma}\right) \\
\color{blue}{E\left(\dfrac{W-\mu}{\sigma}\right)}\ &\color{blue}{= 0}
\end{align*}$
$\underline{\text{To show}\ \small \text{Var}\left(\dfrac{W-\mu}{\sigma} = 1\right):}$
Since we are given that $E(W)=0$ (so that $\mu =0$) and $\text{Var}(W)=1 \lt \infty$, by Theorem 3.6.1 (p. 155)
$\begin{align*}
\text{Var}(W) &= E(W^2) - \mu^2 \\
\sigma^2 &= E(W^2) - 0^2 \quad [\ \text{Theorem 3.6.1}\ ] \\
E(W^2) &= \sigma^2.
\end{align*}$
Thus, $E(W^2) = \sigma^2$ and is finite.
Now, write $\small\dfrac{W-\mu}{\sigma} = \dfrac{1}{\sigma}W + \left(-\dfrac{\mu}{\sigma}\right)$, then use the previous results (the fact that $\mu = 0$ and that $E(W^2) = \sigma^2 \lt \infty$) and Theorem 3.6.2 (p. 156) with $\small a=\dfrac{1}{\sigma}$ and $\small b=-\dfrac{\mu}{\sigma}$ to get
$\begin{align*}
\text{Var}\left(\dfrac{W-\mu}{\sigma}\right) &= \text{Var}\left(\frac{1}{\sigma}W + \left(-\frac{\mu}{\sigma}\right)\right) \\
&= \left(\frac{1}{\sigma}\right)^2\text{Var}(W) \quad [\ \text{Theorem 3.6.2}\ ]\\
&= \frac{1}{\sigma^2}\cdot \sigma^2 \\
\color{blue}{\text{Var}\left(\dfrac{W-\mu}{\sigma}\right)}\ &\color{blue}{= 1}
\end{align*}$
