Answer
(a) $-175^{\circ}C$
(b) $1 \ L$
Work Step by Step
For (a):
From the ideal gas law, $pV = nRT$.
In this case $n, R,$ and $V$ are constant. So we have $\frac{p_1}{T_1} = \frac{p_2}{T_2}$.
$\implies T_2 = T_1(\frac{p_2}{p_1}) = (293)(\frac{1}{3}) = 97.7 K = -175^{\circ}C$.
For (b):
$p_2 = 1 \ atm, V_2 = 3 \ L, p_3 = 3 \ atm$.
In this case, $nRT$ is constant, so we have $p_2 V_2 = p_3 V_3$.
$\implies V_3 = V_2(\frac{p_2}{p_3}) = (3)(\frac{1}{3}) =1 \ L$.