Given,
Current flowing, I = 5 A
Resistance of the resistor, R = 2 Ω
Time, t = 30 min = 1800 s 

Energy transferred = I²Rt
                             = (5)² x 2 x 1800
                             = 9 x 10⁴ J. 

Here, we have
 $$\begin{gathered} \mathrm{Power} \mathrm{of} \mathrm{the} \mathrm{electric} \mathrm{refrigerator}, \mathrm{P} = 400 \mathrm{W} = \frac{400}{1000} \mathrm{kW} = 0.4 \mathrm{kW} \\ Time for which the refrigerator operates, \mathrm{t} = 30 \times 8 \mathrm{h} = 240 \mathrm{h} \end{gathered}$$

$\therefore$ Energy consumed = P x t
                                = 0.4 kW x 240 h
                                = 96 kWh 

Cost of the energy to operate refrigerator at a rate of Rs 3.00 per kWh = 96 x 3.00 = Rs 288.00.

Here,  
Ptential of the generator, V = 220 V
Current flowing across the electric bulb, I = 0.50 A
Therefore,
Power of bulb, P = VI
                         =  220 V x 0.50 A
                         = 110 W. 

Here,
Current flowing through the bulb, I = 4 A
Voltage, V = 12 V
Time, t = 10 min = 600 s 

Energy transferred = Vlt
                             = 12 x 4 x 600
                             = 28800 J