Is my Kirchoff's rules solution with mixed current and voltage source correct?
Assuming your bottom node is assigned the value of \$0\:\textrm{V}\$ and assigning \$V_1\$ to your "1" node and \$V_2\$ to your "2" node, I get:

$$\begin{align*}
\frac{V_1}{R_1} + \frac{V_1}{R_6} + 3 = 0\;\;\;\therefore V_1&amp;=-3\cdot\left(R_1\vert\vert R_6\right)\\&amp;= -25\frac{5}{7}\:\textrm{V}\\\\
\frac{V_2}{R_2+R_3} + \frac{V_2}{R_4} = 3 + \frac{8\:\textrm{V}}{R_2+R_3}\;\;\;\therefore V_2&amp;=\left(3 + \frac{8\:\textrm{V}}{R_2+R_3}\right)\cdot\left(R_4\vert\vert \left[R_2+R_3\right]\right)\\&amp;= 70.\overline{2}\:\textrm{V}
\end{align*}$$

So I get:

$$\begin{align*}
I_4=\frac{V_2}{R_4}&amp;= 1.7\overline{5}\:\textrm{A}\\\\
I_1=\frac{V_1}{R_1}&amp;= -2\frac{4}{7}\:\textrm{A}\\\\
I_2=\frac{8\:\textrm{V}-V_2}{R_2+R_3}&amp;=-1.2\overline{4}\:\textrm{A}\\\\
I_6=\frac{-V_1}{R_6}&amp;=\frac{3}{7}\:\textrm{A}
\end{align*}$$

In short, I think you did just fine.