Total Power Absorbed with KVL and KCL
There is no need for \$i_2\$ since the CCCS in the second branch is causing an integral multiple of \$i_1\$ to flow there and hence the middle \$6k\Omega\$ resistor has \$30+1=31i_1\$ flowing through it.



\$\text{KVL on }M_1:\$
\$
\begin{align}
-5V+(54k\Omega)i_1-1V+(6k\Omega)(31i_1)=0\\
\therefore \quad i_1(54k\Omega+186k\Omega)=6\\
\therefore i_1=\frac{6}{240k\Omega}=25 \mu A \Longleftarrow
\end{align}\$

Voltage across the central \$6k\Omega\$ resistor equals \$6k\Omega \times 31 \times 25\mu A=4.65V\$

Hence  

\$\text{KVL on }M_2:\$
\$
\begin{align}
4.65V-8-(1.8k\Omega\times 30)i_1-\nu=0\\
\therefore \nu=-4.7 V \Longleftarrow
\end{align}\$



Calculations for power dissipation are then easy to take up from this point onwards.