I get different results whenever I try to solve this circuit problem with superposition theorem. Can you please help?
Always re-draw your schematics. And keep in mind that you can select any one node and call it &quot;the ground reference point.&quot; I decided to choose the following arrangement, but you are also free to let your own mind roam and choose differently:



simulate this circuit &ndash; Schematic created using CircuitLab
The first thing that slaps me in the face with the above schematic is that I could treat the current source and \$R_6\$ as a Norton source and immediately convert it into a Thevenin source. This may make things easier. (Though if you don't try such things, you'll never find out.)



simulate this circuit
Simple re-arrangement can make that a little less complex:



simulate this circuit
At this point, you have two voltage sources to deal with and they are both handled the same way. (Note that I've summed two of the resistors in the series loop and called it \$R_\text{SUM}\$.)
Superposition is a broad term that includes two mathematical concepts: additivity and homogeneity. It applies to all linear systems and can apply to vector fields as well as scalar fields.
So when someone tells you to &quot;use superposition&quot; in your answer, they are giving you wide latitude (perhaps without realizing it.) Technically, you could use mesh analysis or nodal analysis, in fact, and I feel successfully argue that you were using superposition.
However, they often mean that you should use the superposition theorem. Which is a more specific thing. The reason I mention all this is that it bugs me that teachers often don't realize that specialists have developed specific terminology for very important reasons and that, at least for those teaching, sloppiness is a vice and not a virtue.
Anyway, assuming you must apply the superposition theorem, then you should be able to readily solve the above. Short out \$V_\text{TH}\$ and then just perform the usual resistor simplifications needed to solve the current: \$I_\text{TOTAL}=\frac{V_\text{TOTAL}=V_g}{R_\text{TOTAL}}=I_{R_1}=I_{R_3}\$. I assume you can work out the value for \$R_\text{TOTAL}\$ in this case, easily. Obviously, this total current leaving \$R_1\$ will divide between \$R_2\$ and the shorted branch, before returning to pass through \$R_3\$. You should be able to work out the portion of the total current found in \$R_2\$. You now have the first part you need in order to solve the problem.
Now, similarly, short out \$V_g\$. All this means is that you tie the top end of \$R_1\$ to ground. Repeat the process now and you can find your second part.
The two steps just discussed look like this:



simulate this circuit
(Note: Please do take note of how freely I re-write the schematics so that I can analyze them more easily. You should do this, too, and do it all the time, every time. Just get into the practice of it. I've been at this for many decades and I still do it and I'll probably never stop doing it. Take this lesson from me deeply into your heart.)
Sum the two relevant currents, \$i_{a_1}\$ and \$i_{b_1}\$, and you have your answer for the current, \$i=I_{R_2}\$.
You will need to look back through your collected information at this point to work out the indicated voltage. But it shouldn't be hard to do, now.
By looking at the third schematic from the top, the &quot;simplified&quot; schematic I mentioned there, you know that the voltage between points c and d are just \$100\:\text{V}\$, less the voltage drop across \$R_\text{TH}\$ (assuming the current direction is outward from the (+) terminal of \$V_\text{TH}\$ and then passing through \$R_\text{TH}\$ on the way out.)
Since you have already all the information you need, \$i_{a_2}\$ (which is pointing the wrong way and therefore must be treated as a negative value for the desired sum) and \$i_{b_\text{TOTAL}}\$ (which is pointing in the right direction for the desired sum), you can work out the net current through \$R_\text{TH}\$ on the way out of the \$100\:\text{V}\$ supply and, with that, compute the drop across \$R_\text{TH}\$ and therefore the voltage between c and d.
That's all there is to the problem.