Answer
See the detailed answer below.
Work Step by Step
a) We have 4 processes, two diabetic and two isochoric processes.
We know that the work done by the system is given by
$$W=W_{1\rightarrow2}+W_{2\rightarrow3}+W_{3\rightarrow4}+W_{4\rightarrow1}$$
For the two isochoric processes, the work done is zero.
$$W=W_{1\rightarrow2}+0+W_{3\rightarrow4}+0$$
In the two adiabatic processes, the heat exchange is zero, so $\Delta E_{th}=-W_s$ thus, $W_s=-\Delta E$
$$W=-nC_{\rm V}(T_2-T_1)-nC_{\rm V}(T_4-T_3)$$
$$W=-nC_{\rm V}[T_2-T_1+T_4-T_3]\tag 1$$
We know that
$$\dfrac{C_{\rm P}}{C_{\rm V}}=\gamma$$
Hence,
$$C_{\rm P} = \gamma\; C_{\rm V}$$
where
$$C_{\rm P}- C_{\rm V} = R$$
Thus,
$$\gamma \;C_{\rm V}- C_{\rm V} = R$$
So
$$C_{\rm V}=\dfrac{R}{\gamma-1}$$
Plugging into (1);
$$W=-\dfrac{nR}{\gamma-1} [T_2-T_1+T_4-T_3] $$
$$\boxed{W= \dfrac{nR}{1-\gamma} [T_2-T_1+T_4-T_3] }$$
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b) We know that the thermal efficiency of the heat engine is given by
$$\eta=\dfrac{W_{out}}{Q_{H}}$$
where $W_{out}=Q_H-Q_C$, thus
$$\eta=1-\dfrac{Q_C}{Q_H}$$
$Q_H$ occurs at the isochoric process from 2 to 3, and $Q_C$ occurs at the isochoric process from 4 to 1.
$$\eta=1-\dfrac{ \color{red}{\bf\not} n \color{red}{\bf\not} C_{\rm V}(T_1-T_4)}{ \color{red}{\bf\not} n \color{red}{\bf\not} C_{\rm V}(T_3-T_2)}$$
$$\eta=1-\dfrac{ (T_1-T_4)}{ (T_3-T_2)}$$
using the ideal gas law $PV=nRT\rightarrow T=\dfrac{PV}{nR}$.
Hence,
$$\eta=1-\dfrac{ P_1V_1-P_4V_4}{P_3V_3-P_2V_2}$$
where $V_1=V_4=V_{\rm max}$ and $V_2=V_3=V_{\rm min}$
$$\eta=1-\dfrac{ P_1V_1-P_4V_4}{P_3V_3-P_2V_2}=1-\dfrac{V_{\rm max}(P_1-P_4)}{V_{\rm min}(P_3-P_2)}$$
where $V_{\rm max}/V_{\rm min}=r$
$$\eta=1-r\left[\dfrac{P_1-P_4}{P_3-P_2}\right] $$
Noting that $P_1\lt P_4$ but $P_3\gt P_2$, so the numerator is negative. and to make it positive,
$$\eta=1+r\left[\dfrac{P_4-P_1}{P_3-P_2}\right]\tag 2$$
where $P_3V_{\rm min}^\gamma=P_4V_{\rm max}^\gamma$, so $P_3=P_4\left[\dfrac{V_{\rm max}}{V_{\rm min}}\right]^\gamma=P_4r^{\gamma}$, and by the same approach, $P_2=P_1\left[\dfrac{V_{\rm max}}{V_{\rm min}}\right]^\gamma=P_1r^{\gamma}$.
Plugging into (2);
$$\eta=1-r\left[\dfrac{P_4-P_1}{P_4r^{\gamma}-P_1r^{\gamma}}\right] $$
$$\eta=1-\dfrac{r}{r^{\gamma}}\left[\dfrac{P_4-P_1}{P_4-P_1}\right] $$
$$\boxed{\eta=1-\dfrac{1}{r^{\gamma-1}} }$$
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c) We know, for diatomic gas, that $C_{\rm V}=\frac{5}{2}R$, and $C_{\rm P}=\frac{7}{2}R$, and hence, $\gamma=7/5$
Plugging into the previous boxed formula above,
$$\eta=1-\dfrac{1}{r^{\frac{7}{5}-1}} $$
$$\boxed{\eta=1-\dfrac{1}{r^{\frac{2}{5}}}} $$
Here is the graph below
