Đáp án:

 $\begin{align}
  & {{R}_{1}}=60\Omega ;{{R}_{2}}=40\Omega  \\ 
 & {{R}_{3}}=16\Omega  \\ 
\end{align}$

Giải thích các bước giải:

 $\begin{align}
  & U=12V;{{R}_{1}}nt{{R}_{2}};{{P}_{nt}}=1,44W; \\ 
 & {{R}_{1}}//{{R}_{2}};{{P}_{//}}=6W \\ 
\end{align}$

khi mắc nối tiếp:

$\begin{align}
  & {{P}_{nt}}=U.{{I}_{nt}}=\dfrac{{{U}^{2}}}{{{R}_{1}}+{{R}_{2}}} \\ 
 & \Rightarrow {{R}_{1}}+{{R}_{2}}=\dfrac{{{12}^{2}}}{1,44}=100\Omega  \\ 
\end{align}$

khi mắc song song: 

$\begin{align}
  & {{P}_{//}}=U.{{I}_{//}}=\dfrac{{{U}^{2}}}{\dfrac{{{R}_{1}}.{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}} \\ 
 & \Rightarrow \dfrac{{{R}_{1}}.{{R}_{2}}}{100}=\dfrac{{{12}^{2}}}{6} \\ 
 & \Rightarrow {{R}_{1}}.{{R}_{2}}=2400(2) \\ 
\end{align}$

từ (1) và (2):

$\begin{align}
  & {{R}_{1}}.(100-{{R}_{1}})=2400 \\ 
 & \Rightarrow {{R}_{2}}=40\Omega ;{{R}_{1}}=60\Omega  \\ 
\end{align}$

b) $\left( {{R}_{1}}//{{R}_{2}} \right)nt{{R}_{3}}$

$\left\{ \begin{align}
  & {{I}_{12}}={{I}_{3}} \\ 
 & U={{U}_{12}}+{{U}_{3}} \\ 
\end{align} \right.$

$\begin{align}
  & \dfrac{{{U}_{12}}}{{{R}_{//}}}=\dfrac{{{U}_{3}}}{{{R}_{3}}} \\ 
 & \Leftrightarrow \dfrac{12-{{U}_{3}}}{24}=\dfrac{{{U}_{3}}}{{{R}_{3}}} \\ 
 & \Leftrightarrow {{R}_{3}}=\dfrac{24{{U}_{3}}}{12-{{U}_{3}}}(3) \\ 
\end{align}$

ta có công suất: 

$\begin{align}
  & {{P}_{3}}=\dfrac{5}{3}{{P}_{1}} \\ 
 & \Leftrightarrow \dfrac{U_{3}^{2}}{{{R}_{3}}}=\frac{5}{3}.\dfrac{{{U}_{//}}^{2}}{{{R}_{1}}} \\ 
 & \frac{U_{3}^{3}}{\dfrac{24{{U}_{3}}}{12-{{U}_{3}}}}=\frac{5}{3}.\dfrac{{{(12-{{U}_{3}})}^{2}}}{60} \\ 
 & \Leftrightarrow \dfrac{{{U}_{3}}}{24}=\dfrac{5}{3}.\frac{12-{{U}_{3}}}{60} \\ 
 & \Rightarrow {{U}_{3}}=4,8V \\ 
\end{align}$

Điện trở R3: 

${{R}_{3}}=\dfrac{24{{U}_{3}}}{12-{{U}_{3}}}=\dfrac{24.4,8}{12-4,8}=16\Omega $