Is a rhombus rigid on a sphere or torus? And generalizations
Q3: Laman's theorem is the same on the sphere.
Indeed, a configuration with $n$ vertices and $m$ edges is defined by a system of $m$ equations in $2n-3$ variables (there are $2n$ coordinates of points, but we may assume that the first point is fixed and the direction of one of the edges from the first point is fixed too). The the left-hand sides are analytic functions of our variables and the right-hand sides are the squares of the lengths of the bars (on the sphere, cosines rather than squares).
Consider this system as a map $f:\mathbb R^{2n-3}\to\mathbb R^m$. The rigidity means that a generic point $x\in\mathbb R^{2n-3}$ cannot be moved within the pre-image of $f(x)$. This implies that $rank(df)=2n-3$ on an open dense set. Choose a configuration from this set and project it to the sphere of radius $R\to\infty$. The equations on the sphere converge to those in the plane, hence the rank of the linearization on the sphere will be maximal ($=2n-3$) for all large $R$. So we get an open set of configuration on the sphere where the linearization has the maximal rank (and this implies rigidity). Since all functions involved are analytic and the rank is maximal on an open set, it is maximal generically. So our linkage is generically rigid on the sphere.
Conversely, consider a flexible linkage on the plane. If $m<2n-3$, it will be flexible on the sphere by a dimension counting argument. Otherwise, by Laman's theorem, there is a subgraph with $r$ vertices and more than $2r-3$ edges. Consider such a subgraph for which $r$ is minimal. Then, by Laman's theorem, we can remove some edges so that this subgraph remains rigid. And, by the above argument, it is rigid on the sphere too. So the edges that we removed were redundant both in the plane and in the sphere. Let's forget about them and repeat the procedure. Eventually we will get a linkage with fewer than $2n-3$ edges.
This might be a cheat, but if you consider a linkage turning once around the meridian of a torus, it might be not flexible. Actually, a linkage homeomorphic to a circle that minimizes length in its homology class is not flexible.
No rhombus on the round sphere is rigid, if we define a rhombus $ABCD$ to be a union of four geodesic segments $\overline{AB}\cup \overline{BC}\cup \overline{CD}\cup \overline{DA}$, such that $A$, $B$, $C$, and $D$ are all distinct points, no three of which are collinear, and such that all four segments have the same length, with self-intersections and overlaps allowed. Suppose $ABCD$ is a rhombus. First note that the hypotheses imply that $B$ is neither equal to nor antipodal to $A$: they're not equal because the vertices are distinct; while if they were antipodal, then $C$ would have to be equal to $A$ because $A$ is the only point at the correct distance from $B$. These facts imply that the endpoints of the geodesic segments starting at $A$ and having length equal to $AB$ trace out a circle, not a point. Let $B'$ be any point on this circle, and let $C'$ be the point obtained by reflecting $A$ through the great circle containing $B'$ and $D$. Because reflection through a great circle is an isometry, it follows that $AB'C'D$ is also a rhombus with the same side lengths as $ABCD$.
I would guess that this argument could be made to work in any symmetric space (with a suitable interpretation of "antipodal"), but I have no idea what would happen with more general metrics.