Đáp án:

 $\begin{align}
  & a){{R}_{1}}=2\Omega ;{{R}_{2}}=4\Omega ;{{R}_{3}}=6\Omega  \\ 
 & b){{U}_{1}}=2V;{{U}_{2}}=4V;{{U}_{3}}=6V \\ 
\end{align}$

Giải thích các bước giải:

$\begin{align}
  & {{R}_{td}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}\Rightarrow {{R}_{td}}=\frac{12}{1}=12\Omega (1) \\ 
 & {{R}_{12}}={{R}_{1}}+{{R}_{2}}\Rightarrow {{R}_{12}}=\frac{12}{2}=6\Omega (2) \\ 
 & {{R}_{13}}={{R}_{1}}+{{R}_{3}}\Rightarrow {{R}_{13}}=\frac{12}{1,5}=8\Omega (3) \\ 
\end{align}$

Từ (1) (2) và (3): $\left\{ \begin{align}
  & {{R}_{1}}+{{R}_{2}}+{{R}_{3}}=12\Omega  \\ 
 & {{R}_{1}}+{{R}_{2}}=6\Omega  \\ 
 & {{R}_{1}}+{{R}_{3}}=8\Omega  \\ 
\end{align} \right.$

$\Rightarrow \left\{ \begin{align}
  & {{R}_{1}}=2\Omega  \\ 
 & {{R}_{2}}=4\Omega  \\ 
 & {{R}_{3}}=6\Omega  \\ 
\end{align} \right.$

b) $\begin{align}
  & {{U}_{1}}=I.{{R}_{1}}=2V \\ 
 & {{U}_{2}}=I.{{R}_{2}}=4V \\ 
 & {{U}_{3}}=I.{{R}_{3}}=6V \\ 
\end{align}$