Hearing the 17 planar symmetry groups Though I'm sure it's not really hard to work out for myself, does anyone know a reference for the spectra of the Laplacian on the 17 flat compact orbifolds that underlie the 17 planar symmetry groups.  
I'm thinking  Neumann boundary conditions to model reflection lines.
And I do realize that these spectra may vary according to a certain number of moduli depending on the group.
(Feel free to add more tags if appropriate.)
 A: I don't think that it's so hard to work out the solution non-rigorously, nor all that onerous to do it rigorously following Ian Agol's suggestion in the comments.  A reference could be nice, but I don't think that it's quite necessary, because the answer itself is not all that different from its derivation.
First, in the Euclidean plane $\mathbb{R}^2$, the plane wave 
$$f(\vec{x}) = \exp(i\vec{k} \cdot \vec{x})$$
is an eigenfunction of the Laplacian $-\nabla$ with eigenvalue $\vec{k} \cdot \vec{k}$.  Suppose that the orbifold is expressed as $X = \mathbb{R}^2/\Gamma$, where $\Gamma$ is a discrete, cocompact group.  Then if you average $f$ with respect to $\Gamma$, you will get a Laplace eigenvector on $X$ with the same eigenvalue.  Since these plane waves are complete (in an analytic sense) upstairs, their averages are at least complete downstairs.
Let $A \subseteq \Gamma$ be the subgroup of translations of $\Gamma$.  Then the $A$-average of $f$ is either $f$ again or it vanishes.  It's $f$ precisely when $\vec{k} \in 2\pi A^*$, where $A^*$ is the dual lattice of $A$.
Then there is a rotation group $R = \Gamma/A$ which is some finite group.  You can now average $f$ over any lift of $R$ and see what you get.  If $R$ lifts to a finite subgroup of $\Gamma$, then that subgroup, call it $R$ again, fixes a point, say the origin.  In this case you get a basis of eigenfunctions using the $R$-orbits of $f$ and $\vec{k}$.  The answer is all $\vec{k} \cdot \vec{k}$, where $\vec{k}$ represents each $R$-equivalence class in $2\pi A^*$.
For example, suppose that $\mathbb{R}^2/A$ is the standard square torus and $A$ is the standard square lattice, so that $A^* = A$.  Suppose that $R$ is generated by a rotation by 90 degrees at the origin.  Then $\vec{k} = 2\pi(n,m)$ and the eigenvalue is $4\pi^2(n^2 + m^2)$, where you should be careful to choose the pair of integers $(n,m)$ with $n \ge 0$ and $m > 0$, or $n = m = 0$.
If $\Gamma$ is a non-split extension of $R$ by $A$, then the answer is a little more complicated, but it's not that much more complicated.  In fact the basic analysis is the same for compact Euclidean orbifolds in any dimension.
