Answer
(a) $V = 274 \,\text{V}$
(b) $P = 1.6 \,\text{W}$
(c) $P_2$ = 2 W, $V$ = 28.8 V and $P_1$ = 1.32 W
Work Step by Step
(a) Using the value of the dissipated power we could get the voltage across as next
\begin{gather*}
P = \dfrac{V^{2}}{R} \\
V = \sqrt{PR} \\
V = \sqrt{(5.0 \,\text{W})( 15000 \mathrm{~ \Omega})} \\
V = 274 \,\text{V}
\end{gather*}
(b) Again we get the dissipated power by
\begin{align*}
P = \dfrac{V^{2}}{R}= \dfrac{(120 \,\text{V})^{2}}{9000 \mathrm{~ \Omega}} = \boxed{1.6 \,\text{W}}
\end{align*}
(c) $R_{1} = 100.0 \mathrm{~ \Omega}$, $R_{2} = 150.0 \mathrm{~ \Omega}$ are in series and the given power is $P = 2\,\text{W}$ for the greater resistor. So, the rated heat due to $R_{2}$ is
$$\boxed{P_{2} = 2\,\text{W}}$$
The current is the same for both resistors and the total voltage is the sum of each resistor. Using the value of $P$ we get the current $I$
\begin{align*}
I = \sqrt{P_{2}/ R_{2}} = \sqrt{(2\,\text{W})/(150.0 \mathrm{~ \Omega})} = 0.115 \,\text{A}
\end{align*}
So, using Ohm's law we get the voltage by
\begin{align*}
V = I(R_{1} + R_{2}) = 0.115 \,\text{A}(100.0 \mathrm{~ \Omega}+ 150.0 \mathrm{~ \Omega} ) = \boxed{28.8 \,\text{V}}
\end{align*}
Let us determine the dissipated power $P_1$ for $R_1$ using the value of $I$ in the form
\begin{align*}
P_{1} = I^{2}R_{1} = (0.115 \,\text{A})^{2} (100.0 \mathrm{~ \Omega}) = \boxed{1.32 \,\text{W}}
\end{align*}
