Answer
Table is on the image
Work Step by Step
a) From table A-6
$P=0.3MPa$
$\upsilon=0.79645\frac{m^3}{kg}$
As $u\gt u_{g}$ is a superheated vapor
b) From table A-5
$T=133.52^{\circ}C$
As $561.11\frac{kJ}{kg}\lt u\lt 2543.2\frac{kJ}{kg}$ is a saturated mixture.
$x=\frac{1560\frac{kJ}{kg}-561.11\frac{kJ}{kg}}{1982.1\frac{kJ}{kg}}=0.504$
$\upsilon=0.001073\frac{m^3}{kg}+0.504*(0.60582\frac{m^3}{kg}-0.001073\frac{m^3}{kg})=0.3059\frac{m^3}{kg}$
c)Insufficient information
d) From table A-4
$u=761.92\frac{kJ}{kg}$
$\upsilon=0.001127\frac{m^3}{kg}$
As $T\lt T_{sat}$ is a compressed liquid
