Answer
a) $ { 4.76}^\circ \rm C$
b) $69.3\;\rm J/kg.K$
Work Step by Step
a)
We need to find the rate of exhausted heat to the river, so we can find the change in temperature.
$$Q_{river}^+=Q_L=mc\Delta T$$
whereas $Q_{river}^+$ is the amount of heat absorbed by the water of the river which is the amount of heat $Q_L$ exhausted from the power plant.
Dividing both sides by $t$;
$$\dfrac{Q_L}{t}=\dfrac{m_{river} c_{river}\Delta T_{river}}{t}$$
We also know, from the density law, that $m=\rho V$
So that,
$$\dfrac{Q_L}{t}=\dfrac{\rho_{river} V_{river} c_{river}\Delta T_{river}}{t}$$
$$ \dfrac{Q_L}{t}=\rho_{river} c_{river}\Delta T_{river}\dfrac{V_{river}}{t}\tag 1$$
And since the author assumed that the turbine operates as an ideal Carnot engine, so
$$e= \dfrac{T_H-T_L}{T_H}=\dfrac{W}{Q_H}=\dfrac{W}{W+Q_L}$$
Thus,
$$ \dfrac{T_H-T_L}{T_H} =\dfrac{W}{W+Q_L}$$
Solving for $Q_L$;
$$T_HW=(T_H-T_L)(W+Q_L)$$
$$W+Q_L=\dfrac{T_HW}{T_H-T_L}$$
$$ Q_L=\dfrac{T_HW}{T_H-T_L}-W=W\left(\dfrac{T_H }{T_H-T_L}-1\right)$$
Dividing both sides by $t$;
$$ \dfrac{Q_L}{t}= \overbrace{ \left[\dfrac{W}{t} \right]}^{ P}\left(\dfrac{T_H }{T_H-T_L}-1\right)\tag 2$$
From (1) and (2);
$$ \rho_{river} c_{river}\Delta T_{river}\dfrac{V_{river}}{t}=P\left(\dfrac{T_H }{T_H-T_L}-1\right) $$
Plugging the known and solving for $\Delta T$;
$$ 1000\cdot 4186\cdot 37 \Delta T_{river} =880 \times 10^6
\left(\dfrac{625}{625-285}-1\right) $$
Thus,
$$\Delta T=\color{red}{\bf 4.76}\;\rm K=\color{red}{\bf 4.76}^\circ C$$
b)
Now we need to find the entropy increase per kilogram of the water of the river.
$$\dfrac{\Delta S}{m}=\dfrac{Q}{T}=\dfrac{mc\Delta T}{mT}$$
Thus,
$$\dfrac{\Delta S}{m}=\dfrac{c\Delta T}{T_{avg}}\tag 3$$
whereas $T_{avg}$ is the average temperature of the water which is given by
$$T_{avg}=\dfrac{T_f+T_i}{2}=\dfrac{T_i+\Delta T+T_i}{2}$$
Plugging the know;
$$T_{avg}=\dfrac{2T_i+\Delta T }{2}=\dfrac{(2\cdot 285)+ 4.76}{2}=\bf 287.4 \;\rm K $$
Plugging into (3) and plug the known as well;
$$\dfrac{\Delta S}{m}=\dfrac{4186\cdot 4.76}{287.4 } $$
$$\dfrac{\Delta S}{m}=\color{red}{\bf 69.3}\;\rm J/kg.K$$
