In a perfectly symmetrical spherical hollow shell, there is a null net gravitational force according to Newton, since in his theory the force is exactly inversely proportional to the square of the distance.
What is the result of general theory of relativity? Is the spacetime flat inside (given the fact that orbit of Mercury rotates I don't think so)? How is signal from the cavity redshifted to an observer at infinity?
Here we will only answer OP's two first question(v1). Yes, Newton's Shell Theorem generalizes to General Relativity as follows. The Birkhoff's Theorem states that a spherically symmetric solution is static, and a (not necessarily thin) vacuum shell (i.e. a region with no mass/matter) corresponds to a radial branch of the Schwarzschild solution
$$\tag{1} ds^2~=~-\left(1-\frac{R}{r}\right)c^2dt^2 + \left(1-\frac{R}{r}\right)^{-1}dr^2 +r^2 d\Omega^2$$
in some radial interval $r \in I:=[r_1, r_2]$. Here the constant $R$ is the Schwarzschild radius, and $d\Omega^2$ denotes the metric of the angular $2$-sphere.
Since there is no mass $M$ at the center of OP's internal hollow region $r \in I:=[0, r_2]$, the Schwarzschild radius $R=\frac{2GM}{c^2}=0$ is zero. Hence the metric (1) in the hollow region is just flat Minkowski space in spherical coordinates.
Recently, it has been shown that an alternative set of solutions exist for the Einstein field equations inside a thin spherical shell. The solutions were initially found by Jun Ni and have since been modified by Lubos Neslusan and Jorge deLyra. In these new solutions, the metric inside a spherical shell is no longer flat and Minkowski. Instead there is a void at the origin and a repulsive gravitational field centred on the origin. A test particle inside the shell will be driven towards the shell. There will also be a redshift seen in light waves emitted by a source in the shell by an observer in the interior.
