At least Q2 is definitely wrong!
Just consider the great dodecahedron. It is non-convex, faces are (convex) regular pentagons, edges are the same ones as those of its covex hull (an icosahedron). But its kernel is a tiny dodecahedron, which touches only onto the faces, but neither onto the edges nor vertices.
The same holds for the great icosahedron, which even has (regular) triangular faces only (as you required). Here the edges are the same as for the stellated dodecahedron, but the kernel is a tiny (convex) icosahedron, which touches only onto the faces, but neither onto the edges nor vertices. (It is only that due to even higher density that the inscribed icosahedron becomes less easy to visualize in this case.)
Edit according to comment discussion:
you well could consider the zigzag push-pleated prism, cf. the also added pic: that one clearly is continously transformable into a fully convex body, i.e. homeomorphic to $S^2$, but still no bit of its edges at B are visible from a point A! - And if you'd contract the lower triangle into a single point (i.e. would use a push-pleated pyramid instead) you get that the point A could see 3 of the edges of P only!
--- rk