Answer
a) $66.7\%$
b) $0.42\;\rm J/K$
c) $0\;\rm J/K$
Work Step by Step
a)
First, we need to find the efficiency of the real heat engine and then for the ideal engine and figure out which is which.
$$e_{real}=\dfrac{W}{Q_H}$$
$$e_{Carnot}=\dfrac{T_H-T_L}{T_H}$$
Divide the real over the ideal:
$$\dfrac{e_{real}}{e_{Carnot}}=\dfrac{\dfrac{W}{Q_H}}{\dfrac{T_H-T_L}{T_H}}$$
Thus,
$$\dfrac{e_{real}}{e_{Carnot}}=\dfrac{W}{Q_H}\cdot \dfrac{T_H}{T_H-T_L} $$
Plugging the known;
$$\dfrac{e_{real}}{e_{Carnot}}=\dfrac{550}{2500}\cdot \dfrac{970}{970-650}=\bf 0.667 $$
Thus,
$$\dfrac{e_{real}}{e_{Carnot}} = 0.667\times 100\%=\color{red}{\bf 66.7}\% $$
b)
The temperature of the heat reservoirs does not change during the operation.
But the input reservoir loses heat and hence there is an entropy loss. And the output reservoir gains heat and hence there is an entropy gaining.
Thus,
$$\Delta S=\Delta S_{loss}+\Delta S_{gaining}=\dfrac{-Q_H}{T_H}+\dfrac{Q_L}{T_L}$$
We know that $Q_H=Q_L+W$, and hence, $Q_L=Q_H-W$.
$$\Delta S =\dfrac{-Q_H}{T_H}+\dfrac{Q_H-W}{T_L}$$
Plugging the known;
$$\Delta S =\dfrac{-2500}{970}+\dfrac{2500-550}{650}$$
$$\Delta S =\color{red}{\bf 0.42}\;\rm J/K$$
c)
For the Carnot engine:
$$\Delta S=\Delta S_{loss}+\Delta S_{gaining}=\dfrac{-Q_H}{T_H}+\dfrac{Q_L}{T_L}\tag 1$$
We know, for Carnot engine, that
$$Q_H= \dfrac{Q_L}{1-e_{Carnot}}=\dfrac{Q_L}{1-\left(1-\dfrac{T_L}{T_H}\right)}=\dfrac{Q_LT_H}{T_L}$$
Hence, $Q_L=\dfrac{Q_HT_L}{T_H}$.
plugging that into (1);
$$\Delta S =\dfrac{-Q_H}{T_H}+\dfrac{\dfrac{Q_HT_L}{T_H}}{T_L}=\dfrac{-Q_H}{T_H}+\dfrac{Q_H}{T_H}$$
Therefore,
$$\Delta S =\color{red}{\bf 0}\;\rm J/K$$
