Is spacetime-curvature relative? Velocity is relative, which means kinetic energy is. Since, according to general relativity, energy bends spacetime around it, wouldn't this mean observers moving in different inertial frames measure different values of curvature?
I think this is slightly different from this question: Is space curvature relative?
 A: It depends on what you mean by “curvature”.
The most complete description of curvature is in terms of something called the Riemann curvature tensor, $R_{\mu\nu\lambda\kappa}$. In four dimensions, it has 256 components (only 20 of which are independent), and the value of these components is different in different reference frames. There is a transformation rule for how this tensor transforms between frames. In other words, the components of the Riemann curvature tensor are relative, i.e. observer-dependent.
However, from this tensor you can construct invariants such as the Ricci scalar $R$ which have the same value in all reference frames. These curvature invariants are absolute, i.e., observer-independent.
This is similar to how energy and momentum are relative — they are components of a four-vector which transforms between frames — but a particular combination of them, mass, is absolute.
A: No, it is absolute. The presence of curvature (non vanishing Riemann tensor) is mathematically equivalent to the existence of geodesical deviation of timelike geodesics. This physically means that there is a  relative acceleration of free falling bodies. This fact is absolute, it does not depend on any reference frame as it is a relative phenomenon.
A: Locally, yes. Globally no. 
The Equivalence principle tells you that you can always find a local Inertial reference frame, which means that given a point in spacetime you can always find neighborhood of this point where spacetime is flat, and special relativity holds. 
Globally you can't do that because spacetime is, mathematically speaking, a Riemannian manifold, and its curvature is an intrinsic property. 
