#### Answer

a)$W = -44 kJ$
$W = -55kJ$
c)$T = 151.83°C$

#### Work Step by Step

a) At the first state, the steam at $1\ MPa$ and $400°C$ has a specific volume of $v_1 = 0.30661\ m³/kg$, therefore, given its mass of $0.6\ kg$:
$V_1 = 0.184 m³$
Similarly at the final state, $1\ MPa$ and $250°C$ $v_2 = 0.23275\ m³/kg$:
$V_2 = 0.140 m³$
The work from isobaric cooling is $W = P(V_2-V_1)$
$W = -44 kJ$
b) With the final volume being 40% of the original volume:
$V_2 = 0.074 m³$
$W = P(V_2-V_1)$
$W = -55kJ$
c)The specific volume in this final state is $v_2 = 0.1227\ m³/kg$
By the steam tables, since this value is lower than saturated steam and bigger than saturated water the final state is saturated with part of the steam already condensed: $T = 151.83°C$