I will start from the first bit for my reference
a) P= I2R
P=VI
1500=240XI
I=6.25A
1500=(6.25)2R
R=38.4 ohms
b)
1) When S1 is open, the circuit is not a closed one and hence not functional. Power will thus be 0
2) When only S3 is open, Current flows through A and wire with S2...power of each heating element is given as 1.5kW(stated in the question)...so in this case it is that of A. Current doesn't flow through B because it offers more resistance than wire with S2.
3)when all switches r closed...No current flows through B. You must know that any charge will take the path with least resistance....
So when S2 is closed, current flows through the wire with S2 since it offers a lesser resistance in comparison with wire with heating element B. So Power would be that of A nd C= 1.5+1.5=3kW
4)When only S1 is closed, Current flows through A and B...combined Resistance is 76.8ohms...I=3.125A..P=I2R =0.75kW
5)When only S2 is open, current flows through all the heating elements A, B and C.
A and B in series= 76.8ohms
With C in parallel
So combined R
1/R= 1/76.8+1/38.4
R=25.6
I=240/25.6=9.375
P=9.3752X25.6=2.25kW
Hope that clears ur problem