Answer
a) $0.088$
b) $0.5$
Work Step by Step
$\textbf{In the first process}[1\rightarrow 2],$ which is an isothermal one, the change in the thermal energy is zero.
So,
$$\boxed{(\Delta E_{th})_{1\rightarrow 2}= \color{red}{\bf 0}\;\rm J} $$
Thus,
$$(W_s)_{1\rightarrow 2}=Q_{1\rightarrow 2}=\int_1^2PdV=nRT_1\int_1^2\dfrac{dV}{V}$$
$$(W_s)_{1\rightarrow 2}=Q_{1\rightarrow 2}=nRT_1\ln\left[\dfrac{V_2}{V_1}\right]$$
where the volume is halved.
$$(W_s)_{1\rightarrow 2}=Q_{1\rightarrow 2}=nRT_1\ln\left[\dfrac{\frac{1}{2}V_1}{V_1}\right]$$
$$\boxed{(W_s)_{1\rightarrow 2}=Q_{1\rightarrow 2}=-\ln(2)nRT_1}$$
Since this is an isothermal process $\boxed{T_1=T_2}$, so
$$P_1V_1=P_2V_2=P_2(\frac{1}{2}V_1)$$
where $\boxed{V_2=\frac{1}{2}V_1}$
Thus,
$$\boxed{P_2=2P_1}$$
$\textbf{In the second process}[2\rightarrow 3],$ which is an isobaric one,
$$(W_s)_{2\rightarrow 3}=P_2\Delta V $$
where $P_2=2P_1$, $V_3=V_1$ and $V_2=\frac{1}{2}V_1$
Hence,
$$(W_s)_{2\rightarrow 3}= 2P_1(V_1-\frac{1}{2}V_1)$$
$$\boxed{(W_s)_{2\rightarrow 3}= P_1V_1=nRT_1}$$
$$(\Delta E_{th})_{2\rightarrow 3}=nC_{\rm V}\Delta T=\frac{5}{2}nR(T_3-T_2)$$
And,
$$Q_{2\rightarrow 3}=nC_{\rm P}\Delta T=\frac{7}{2}nR(T_3-T_2)$$
And since it is an isobaric process, so
$$\dfrac{V_2}{T_2}=\dfrac{V_3}{T_3}$$
where $V_2=\frac{1}{2}V_1$, $V_3=V_1$, and $T_2=T_1$
$$T_3=\dfrac{T_2V_3}{V_2}=\dfrac{V_1T_1}{\frac{1}{2}V_1}=
2T_1$$
$$\boxed{T_3 =2T_1}$$
Plugging into
$$(\Delta E_{th})_{2\rightarrow 3}= \frac{5}{2}nR(2T_1-T_1)$$
$$\boxed{(\Delta E_{th})_{2\rightarrow 3}=\frac{5}{2}nRT_1}$$
And hence,
$$\boxed{Q_{2\rightarrow 3}=\frac{7}{2}nRT_1}$$
$\textbf{In the third process}[3\rightarrow 1],$ which is an isochoric one,
$$\boxed{(W_s)_{3\rightarrow 1}=0\;\rm J} $$
Hence,
$$(\Delta E_{th})_{3\rightarrow1}= Q_{3\rightarrow1}=\frac{5}{2}nR(T_1-T_3)=\frac{5}{2}nR(T_1-2T_1)$$
$$\boxed{(\Delta E_{th})_{3\rightarrow1}= Q_{3\rightarrow1}=-\frac{5}{2}nRT_1 }$$
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a) The thermal efficiency of this heat engine is given by
$$\eta=\dfrac{W_{net}}{Q_{in}}=\dfrac{(W_s)_{1\rightarrow 2}+(W_s)_{2\rightarrow 3}+(W_s)_{3\rightarrow 1}}{Q_{2\rightarrow3}}$$
$$\eta=\dfrac{-\ln(2)nRT_1+nRT_1+0}{\frac{7}{2}nRT_1}=\dfrac{-\ln(2)+1}{\frac{7}{2}}$$
$$\eta=\color{red}{\bf 0.088}$$
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b) The maximum thermal efficiency of this heat engine is given by
$$\eta_{\rm max}=1-\dfrac{T_{\rm min}}{T_{\rm max}}=1-\dfrac{T_1}{2T_1}$$
$$\eta_{\rm max}=\color{red}{\bf 0.5}$$
