finding phase and group delay from frequency response
Define a frequency response

$$\tilde{H}(e^{j\omega})=\frac{e^{-j\omega}-|\alpha|}{1-|\alpha|e^{-j\omega}}\tag{1}$$

and note that the group delay $\tilde{\tau}(\omega)$ corresponding to $(1)$ is related to the group delay $\tau(\omega)$ of the original frequency response $H(e^{j\omega})$ by

$$\tau(\omega)=\tilde{\tau}(\omega-\theta)\tag{2}$$

because $(1)$ is just a scaled and frequency-shifted version of the original $H(e^{j\omega})$.

$\tilde{H}(e^{j\omega})$ is the frequency response of a first-order all-pass filter with a real-valued pole at $z_{\infty}=|\alpha|&lt;1$, and a real-valued zero at $z_0=1/|\alpha|$. The original frequency response $H(e^{j\omega})$ describes a first-order all-pass filter with a complex-valued pole at $z_{\infty}=\alpha=|\alpha|e^{j\theta}$ ($|\alpha|&lt;1$), and a zero at $z_0=1/\alpha^*$.

Now write

$$\tilde{H}(e^{j\omega})=\frac{B(e^{j\omega})}{A(e^{j\omega})}\tag{3}$$

and note that

$$\tilde{\phi}(\omega)=\phi_B(\omega)-\phi_A(\omega)\tag{4}$$

where $\tilde{\phi}(\omega)$ is the phase of $\tilde{H}(e^{j\omega})$, and $\phi_A(\omega)$ and $\phi_B(\omega)$ are the phases of $A(e^{j\omega})$ and $B(e^{j\omega})$, respectively. Consequently,

$$\tilde{\tau}(\omega)=\tau_B(\omega)-\tau_A(\omega)\tag{5}$$

where $\tau_A(\omega)$ and $\tau_B(\omega)$ are the group delays of $A(e^{j\omega})$ and $B(e^{j\omega})$, respectively.

We have

$$\phi_B(\omega)=-\arctan\left(\frac{\sin(\omega)}{\cos(\omega)-|\alpha|}\right)\tag{6}$$

and

$$\phi_A(\omega)=\arctan\left(\frac{|\alpha|\sin(\omega)}{1-|\alpha|\cos(\omega)}\right)\tag{7}$$

where I've ignored any ambiguities by $\pm\pi$ due to the principal value of $\arctan()$ because they are irrelevant for computing the group delay.

Taking the negative derivative of $(6)$ and $(7)$ with respect to $\omega$ gives

$$\tau_B(\omega)=\frac{1-|\alpha|\cos(\omega)}{1-2|\alpha|\cos(\omega)+|\alpha|^2}\tag{8}$$

and

$$\tau_A(\omega)=\frac{|\alpha|^2-|\alpha|\cos(\omega)}{1-2|\alpha|\cos(\omega)+|\alpha|^2}\tag{9}$$

From $(5)$ we get

$$\tilde{\tau}(\omega)=\frac{1-|\alpha|^2}{1-2|\alpha|\cos(\omega)+|\alpha|^2}=\frac{1-|\alpha|^2}{\big|1-|\alpha|e^{-j\omega}\big|^2}\tag{10}$$

From $(2)$ the group delay of the complex first-order all-pass is given by

$$\tau(\omega)=\frac{1-|\alpha|^2}{1-2|\alpha|\cos(\omega-\theta)+|\alpha|^2}=\frac{1-|\alpha|^2}{\big|1-|\alpha|e^{-j(\omega-\theta)}\big|^2}\tag{11}$$

The difference with the given solution in Schaum's outline is the fact that in $(11)$ you have $|\alpha|$ instead of the complex-valued $\alpha$. Note that $(11)$ is correct and Schaum's formula is wrong because a complex-valued group delay doesn't make much sense.