Calculating current and voltage using superposition
You could certainly use superposition to solve this problem, but I think that node voltage would probably be much easier to use in this case due to the fact that you have only two unknown voltages in the circuit, but even then you have a relationship between the unknown voltages (due to the voltage source connected to either unknown node) which makes your life much easier. The voltage at the other node (call it v1), can be found as a function of vo due to the fact that, since a voltage source is connected to both of these nodes (call it vs),we can subtract the voltage at v1 from the voltage at vo to determine the vs. Also this can be written as:$$ v_s = v_o-v_1 $$ Therefore: $$ v_1=v_o-v_s $$ 
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Also you said that you used node voltage already, but you aren't sure how that worked. Does this mean that you solved it correctly and are unsure of the philosophy behind the node voltage method or you are unsure of your answer?$$ $$
And to answer your question about the imaginary numbers: you calculate the node voltage using complex numbers (which is a combination of both real and imaginary parts). For example, to find the current in the branch of the circuit with the inductor and resistor (call it i1), you would simply write: $$ i_1 = \frac{v_o}{Z_1}=\frac{v_o}{1+j}   $$ Where Z1 is called the complex impedance of the branch, which is essentially the same as resistance, but also phase shifts the current (in case you haven't already covered complex impedance). 


Let me know if you have any questions!