Explanation

(a) The efficiency of heat engine is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta = \frac{W}{Q} = 1 - \frac{Q_{2}}{Q_{1}} = 1 - \frac{T_{2}}{T_{1}}$
where $T_{1}$ is temperature of source and $T_{2}$ is temperature of sink.
Given, $\eta_{1} = \frac{1}{6},\ \ \ \eta_{2} = \frac{1}{3}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{6} = \frac{T_{1} - T_{2}}{T_{1}}$ …(i)
and $\frac{1}{3} = \frac{T_{1} - (T_{2} - 62)}{T_{1}}$ …(ii)
Solving Eqs. (i) and (ii), we get
${\ \ \ \ \ \ \ \ \ \ T}_{1} = 372\ K$
and $T_{2} = 310\ K$