In certain cases it's true, in others its not. The proof for manifolds with boundary follows very much in the same spirit as the proof for ones without, but there are a few extra complicating details.
It helps to think about the case where it fails. For example, consider embeddings of a compact manifold with boundary $W$ in another compact manifold with boundary $M$. Any example works, but take $W=[0,1]$ and $M=[0,1]$. For such embeddings the isotopy extension theorem does not hold. This fibration theorem you are interested in is just the isotopy extension theorem "with parameters" so it clearly fails.
The reason for the failure: In the proof you need to extend a vector field on the image of $W$ to a vector field with prescribed behaviour outside of a small neighbourhood of the image of $W$. This is generally impossible when both manifolds have boundary -- in this case because the complement can be empty.
So the theorem is true if both $W$ and $M$ have boundary but you require the embeddings to be proper meaning they send boundary to boundary (and being transverse to $\partial M$).
The theorem is also true if $W$ has boundary but $M$ does not. The theorem is also true if $W$ has no boundary, and $M$ has boundary, but you require the embedding to be disjoint from $\partial M$. etc.
I suggest reading the proof of the isotopy extension theorem in Hirsch's Differential Topology text. That proof gives you the basic idea, even for the "with parameters" versions of the theorem that you are interested in.
Regarding (1), for example, Cerf's dissertation made heavy usage of these facts. It is a large IHES publication and I believe it has all these results in it, with fairly detailed proofs. Most publications that use these results have sketches of them. The Goodwillie-Weiss embedding calculus, for example, makes heavy usage of these theorems.
Regarding (2), yes it fails for $M$ having boundary precisely in the instances I described above.
When $V$ is non-compact it can fail as well. For example, $Emb(\mathbb R, \mathbb R^3)$ is pathwise connected, so if the the map $Emb(\mathbb R^3 , \mathbb R^3) \to Emb(\mathbb R, \mathbb R^3)$ were a fibration there would be an ambient isotopy between a "long trefoil" and an unknot.