We will abbreviate by removing $x_0$ from $f(x_0), \hat{f}(x_0), G(x_0)$, and $\hat{G}(x_0)$.
Show (7.62)
Let $\hat{g} = Pr(\hat{f} > \frac{1}{2})$. Then
\[Err(x_0) = E[I[Y \neq \hat{G}] \lvert X = x_0] = Pr(Y \neq \hat{G} \lvert X = x_0) = f (1 - \hat{g}) + (1-f) \hat{g} = f - 2 f \hat{g} + \hat{g}\] \[Err_B(x_0) + \lvert 2f - 1 \rvert P(\hat{G} \neq G \lvert X = x_0) = \begin{cases} f + (1 - 2f) \hat{g} & \text{if } f \leq \frac{1}{2} \\ (1 - f) + (2f - 1) (1 - \hat{g}) & \text{else} \end{cases} = f - 2 f \hat{g} + \hat{g}\]Show (7.63)
\[Pr(\hat{G} \neq G \lvert X = x_0) = Pr(\hat{f} > \frac{1}{2})Pr(f \leq \frac{1}{2}) + Pr(\hat{f} \leq \frac{1}{2})Pr(f > \frac{1}{2})\]If $f > \frac{1}{2}$, above reduces to $Pr(\hat{f} \leq \frac{1}{2})$. Since $\frac{\hat{f} - E[\hat{f}]}{\sqrt{Var[\hat{f}]}} \sim N(0,1)$, we obtain $Pr(\hat{f} \leq \frac{1}{2}) = \Phi(\frac{\frac{1}{2} - E[\hat{f}]}{\sqrt{Var[\hat{f}]}})$. Then, (7.63) holds because sign($\frac{1}{2} - f$) is negative. Similar (sign-reversed) argument holds for the $f \leq \frac{1}{2}$ case.