Solve for the current in the following circuit using superposition
The superposition theorem just says to replace remaining voltage sources with shorts and current sources with opens and evaluate. Then just sum everything up. I don't think any of your results are correct, yours or the ones you think may be right. But perhaps I just can't read your problem with understanding. I can read the circuit, though. Just to recall, here it is:







simulate this circuit &ndash; Schematic created using CircuitLab

Let's focus on \$V_1\$, shorting out \$V_2\$ to produce this for step 1:







simulate this circuit

We can now lay out these currents:

\$I_{R_1} = \frac{V_1}{R_1}\$

\$I_{R_2} = \frac{V_1}{R_2}\$

\$I_{R_3} = 0\$

Note that \$R_3\$ is shorted out by the replaced \$V_2\$ and so its current must be zero.

Now let's focus on \$V_2\$, shorting out \$V_1\$ to produce this for step 2:







simulate this circuit

We can now lay out these currents:

\$I_{R_1} = 0\$

\$I_{R_2} = -\frac{V_2}{R_2}\$

\$I_{R_3} = \frac{V_2}{R_3}\$

Note that \$R_1\$ is shorted out by the replaced \$V_1\$ and so its current must be zero. Also notice that the direction of the current in \$R_2\$ is the opposite to the earlier direction. So here we use an opposing sign. Just be consistent about this.

Now we can simply take the two above cases and sum them up together as though they are happening simultaneously:

\$I_{R_1} = \frac{V_1}{R_1} + 0=\frac{V_1}{R_1}\$

\$I_{R_2} = \frac{V_1}{R_2}-\frac{V_2}{R_2}= \frac{V_1-V_2}{R_2}\$

\$I_{R_3} = 0 + \frac{V_2}{R_3}=\frac{V_2}{R_3}\$

The current in \$V_1\$ will be the sum of the two returning currents from \$R_1\$ and \$R_2\$ or else it will be the sum of the currents through \$R_1\$, \$V_2\$, and \$R_3\$, depending on which way you'd rather look. Either way, it has to be the same.

Just by way of making sense of things, it should be clear that since \$V_1\$ is directly across \$R_1\$ that the final summed current through \$R_1\$ must simply be \$\frac{V_1}{R_1}\$. And it is. Good. Similarly, it should be clear that since \$V_2\$ is directly across \$R_3\$ that the final summed current through \$R_3\$ must simply be \$\frac{V_2}{R_3}\$. And it is. Also good.

Back to business. Easiest way to get \$I_{V_1}\$ is to sum the returning currents in \$R_1\$ and \$R_2\$:

\$I_{V_1}= I_{R_1}+I_{R_2}= \frac{V_1}{R_1}+\frac{V_1-V_2}{R_2}\$

I don't recall seeing that answer in the stuff you provided. Maybe I just didn't see it.