I am reasonably certain this is the case, but can't find a reference that actually states this, although the Wikipedia article states something close.
It is possible to have all the subsets of R be measurable (Solovay, Robert M. (1970). "A model of settheory in which every set of reals is Lebesgue measurable". Annals of Mathematics. Second Series 92: 1–56.) which implies the nonexistence of a well ordering of R. 


Yes. Here's a sketched example: Start in L. Let P be the forcing which adds ω_{1} many Cohen reals, and let G be an Lgeneric filter for P. Then L(ℝ)^{L[G]} will model ZF, but will have no well ordering of the reals. The point is that if σ is an automorphism of P, then σ can be extended to an elementary map from L[G] to L[σ[G]], and this extension will fix L(ℝ)^{L[G]}. So if there was a well ordering of ℝ in L(ℝ)^{L[G]}, it would give a well ordering of G which was fixed by σ. But σ can reorder the elements of G because of the homogeneity of P. 

