Answer
(a) less than
(b) $0.46$
(c) $-1.8J/K$
(d) $1.8J/K$
Work Step by Step
(a) For the first case
$e=1-\frac{T_c}{T_h}$
$\implies e=1-\frac{305K}{576K}$
$e=0.47$
For the new case
$e=1-\frac{T_c}{T_h}$
$e=1-\frac{321K}{592K}$
$\implies e=0.4577$
Thus, we can see that the new efficiency is less than $0.47$.
(b) We know that
$e=1-\frac{T_c}{T_h}$
$\implies e=1-\frac{321K}{592K}$
$e=0.46$
(c) As $\Delta S_h=-\frac{Q_h}{T_h}$
We plug in the known values to obtain:
$\Delta S_h=-\frac{1050J}{592K}$
$\Delta S_h=-1.8J/K$
(d) We know that
$\Delta S_c=\frac{Q_h(1-e_{max})}{T_c}$
We plug in the known values to obtain:
$\Delta S_c=\frac{1050J(1-0.4577)}{321K}$
$\Delta S_c=1.8J/K$
