Đáp án:

$\begin{array}{l}
a.{R_1} = 90\left( \Omega  \right)\\
{R_2} = 60\left( \Omega  \right)
\end{array}$

b.đèn 1 cháy

đèn 2 sáng yếu

c.$\begin{array}{l}
\left( {{R_1}//R} \right)nt{R_2}\\
R = 180\left( \Omega  \right)
\end{array}$

Giải thích các bước giải:

$\begin{array}{l}
a.{R_1} = \frac{{U_{dm1}^2}}{{{P_{dm1}}}} = \frac{{{{30}^2}}}{{10}} = 90\left( \Omega  \right)\\
{R_2} = \frac{{U_{dm2}^2}}{{{P_{dm2}}}} = \frac{{{{30}^2}}}{{15}} = 60\left( \Omega  \right)\\
b.{R_1}nt{R_2}\\
{R_{td}} = {R_1} + {R_2} = 150\left( \Omega  \right)\\
I = \frac{U}{{{R_{td}}}} = \frac{{60}}{{150}} = 0,4\left( A \right)\\
{U_1} = I.{R_1} = 0,4.90 = 36 > {U_{dm1}}
\end{array}$

⇒đèn 1 cháy

${U_2} = I.{R_2} = 0,4.60 = 24 < {U_{dm2}}$

⇒đèn 2 sáng yếu

c. phải mắc 

$\begin{array}{l}
\left( {{R_1}//R} \right)nt{R_2}\\
{I_{1R}} = {I_2}\\
 \Rightarrow \frac{{{U_{dm1}}}}{{{R_{1R}}}} = \frac{{{U_{dm2}}}}{{{R_2}}}\left( {{U_{dm1}} = {U_{dm2}} = 30} \right)\\
 \Rightarrow {R_2} = \frac{{{R_1}R}}{{{R_1} + R}}\\
 \Rightarrow 60 = \frac{{90R}}{{90 + R}}\\
 \Rightarrow R = 180\left( \Omega  \right)
\end{array}$