Question

[Key Concept: Parallel branches share the same voltage; the series resistor carries the total current.]

in parallel $I \propto \frac{1}{R}$ so current

in $5\Omega$ will be 3 times that of current in $15\Omega$.

$$\begin{matrix} & P_{5\Omega} = (3I)^{2} \times 5 \\ & 7.2 = 9 \times 5 \times I^{2} \\ & I^{2} = \frac{7.2}{9 \times 5} = \frac{0.8}{5} = 0.16 \\ & I = 0.4\text{ }A \\ & P_{6\Omega} = I^{2} \times 6\Omega = 0.16 \times 6 = 0.96\text{ }W \end{matrix}$$

Statement (1) is wrong.

$$\begin{matrix} V_{AB} & \ = 4 \times 4I + 5 \times 3I \\ & \ = 31I \\ & \ = 31 \times 0.4\text{ }V \\ & \ = 12.4Volt \end{matrix}$$

Statement (2) is correct