Why is the cis isomer of 1,3-dimethylcyclobutane more stable than the trans isomer?
The four carbon atoms in cyclobutane are not coplanar, as that would lead to a large degree of unfavourable eclipsing interactions. So, it is not a flat square; it adopts a puckered conformation. One of the carbon atoms makes a $25^\circ$ angle with the plane formed by the other three carbons, and this mitigates some of the eclipsing interactions, at the cost of a small increase in angle strain. The conformation is commonly called the "butterfly", and two equivalent puckered conformations interconvert rapidly. Here's a visual reference to help you 



Pictured above is an unsubstituted cyclobutane.

Take a closer look at the figure on the left, and notice how there would be some transannular interaction between the two $\ce{H'}$ hydrogens at $\ce{C-1}$ and $\ce{C-3}$, marked in $\color{red}{\text{red}}$. This is bad. In the figure on the right, the $\ce{H'}$ hydrogens are no longer axial, but equatorial, and this is better. 

In short, placing substituents in an equatorial position is better than placing them in an axial position.

Now, imagine what the cis and trans isomers would look like for 1,3-dimethylcyclobutane. 

In the cis isomer both $\ce{H'}$ on $\ce{C-1}$ and $\ce{C-3}$ would be replaced by $\ce{-CH3}$ groups (since they are on the same side). Of course there will be considerable repulsions if they are arranged in the conformation on the left, which is why ring flip takes place and we will predominantly get the conformation on the right with both the methyl groups in equatorial positions. This is ideal.



The trans isomer would be like replacing one $\color{red}{\text{red}}$ $\ce{H'}$ and one $\color{blue}{\text{blue}}$ $\ce{H}$ at $\ce{C-1}$ and $\ce{C-3}$ with methyl groups. No matter what you do, you can only get one of the methyl groups at the equatorial position, but never both.