Solution

Here the isothermal process proceeds at the maximum temperature instead of at the minimum temperature of the cycle.
(a) Here $$p_1 V_1 = p_0V_0, p_2 = \frac{p_1}{n}$$
$$p_2V_2^{\gamma} = p_0V_0^{\gamma}$$ or $$ p_1V_1^{\gamma} = np_0 V_0^{\gamma}$$
i.e., $$V_1^{\gamma -1} = nV_0^{\gamma -1}$$ or $$V_1 = V_0 n^{\frac{1}{\gamma -1}}$$
$$Q_2' = C_VT_0\Big( 1 - \frac{1}{n}\Big), Q_1 = RT_0 In \frac{V_1}{V_0}= \frac{RT_0}{\gamma - 1}In \ n = C_V T_0 In \ n$$.
Thus $$\eta = 1 - \frac{Q_2'}{Q_1} = 1 - \frac{n-1}{n \ In \ n}$$
(b) Here $$ V_2 = \frac{V_1}{n} , \ p_0V_0 = p_1V_1$$
$$p_0V_0^{\gamma} = p_1V_2^{\gamma} = p_1 n^{-\gamma}V_1^{\gamma} =V_0^{\gamma -1} n^{-\gamma} V_1^{\gamma -1}$$ or $$V_1 = n^{(\gamma / \gamma -1)} V_0$$
$$Q_2' = C_pT_0\Big(1 - frac{1}{n}\Big)$$, $$Q_1 = RT_0 \ In \ \frac{V_1}{V_0} = \frac{R\gamma}{\gamma -1} T_0 \ In \ n = C_pT_0 \ In \ n$$
Thus $$\eta = 1 - \frac{n-1}{n \ In \ n}$$