Answer
(a)
Use KCL at Node A.
$$
\begin{array}{r} {\quad i_{1}=6 A}\end{array}
$$
---
(b)
Use KCL at Node B.
$$
\begin{aligned} i_{2} &=-4 A \end{aligned}
$$
Work Step by Step
(a)
Use KCL at Node A.
$$
\begin{array}{r}{i_{1}+i_{b}-i_{a}-i_{c}=0} \\ {\qquad i_{1}=i_{a}-i_{b}+i_{c}} \\ {\quad i_{1}=6 A}\end{array}
$$
---
(b)
Use KCL at Node B.
$$
\begin{aligned} i_{e}+i_{d}-i_{1}-i_{2}=0 \\
\begin{aligned} i_{2} &=i_{e}+i_{d}-i_{1} \\ i_{2} &=-4 A \end{aligned}
\end{aligned}
$$
