Carnot Engine

Efficiency of the engine: $\eta = 1 - \dfrac{T_2}{T_1}$, where $T_1$ is the temperature of the source and $T_2$ that of the sink (in kelvin).
According to the problem, $\dfrac{1}{6} = 1 - \dfrac{T_2}{T_1}$ ...(1)
From (1): $\dfrac{T_2}{T_1} = \dfrac{5}{6}$
When the temperature of the sink is reduced by $62^\circ C$ (a fall of 62 K), the efficiency is doubled: $2\left(\dfrac{1}{6}\right) = 1 - \dfrac{T_2 - 62}{T_1}$ ...(2)
$\dfrac{1}{3} = 1 - \dfrac{T_2}{T_1} + \dfrac{62}{T_1} = \dfrac{1}{6} + \dfrac{62}{T_1}$
$\dfrac{62}{T_1} = \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{1}{6}$
$T_1 = 62 \times 6 = 372$ K
$T_1 = 372 - 273 = 99^\circ C$ = temperature of the source.
(The sink is at $T_2 = \dfrac{5}{6} \times 372 = 310$ K $= 37^\circ C$.)

Efficiency of a Carnot engine: $\eta = 1 - \dfrac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ the sink temperature.
$1 - \dfrac{300}{T_1} = 0.4 \Rightarrow \dfrac{300}{T_1} = 0.6 \Rightarrow T_1 = 500$ K
The efficiency is to be increased by 50 percent of the original efficiency: $\eta' = 0.4 + 0.4 \times \dfrac{50}{100} = 0.6$
Let the source temperature be increased by $\Delta T$, the sink remaining at 300 K.
$0.6 = 1 - \dfrac{300}{500 + \Delta T}$
$\dfrac{300}{500 + \Delta T} = 0.4 \Rightarrow 500 + \Delta T = 750$
$\Rightarrow \Delta T = 250$ K

$T_1 = 227 + 273 = 500$ K (source) and $T_2 = 127 + 273 = 400$ K (sink)
Efficiency: $\eta = \dfrac{W}{Q_1} = 1 - \dfrac{T_2}{T_1}$
$\eta = 1 - \dfrac{400}{500} = 0.2$
$W = Q_1\left(1 - \dfrac{T_2}{T_1}\right) = 6 \times 10^{4} \times 0.2$
$W = 1.2 \times 10^{4}$ cals