For details I recommend looking at the papers of Yagasaki on arXiv
especially the paper

Let me answer 1. Consider be the diffeomorphism group of $\mathbb R^2$ that fixes $0$ and acts as the identity at the tangent space to $0$. The group acts on your space of embeddings by postcomposition. By the parametrized isotopy extension theorem the action is transitive on each path-component and the orbit map is a fiber bundle whose fiber is the subspace of diffeomorphisms that are identity on the inclusion of the disk. The latter space is contractible by the Alexander trick towards infinity. Note that so far the two-dimensionality has not been used and the argument generalizes to the embedding of a compact manifold onto an open manifold obtained by attaching a collar to the compact one.

We are left to understand the homotopy type of the above diffeomorphism group. The group is contractible as proved in the last mentioned paper of Yagasaki (or even easier, look at the Anton Petrunin's answer which really proves contractibility of the diffeomorphism group). Thus each component of the embedding space is contractible.
In fact, the embedding space is path-connected essentially because any two embedded circles in $\mathbb R^2-\{0\}$ are isotopic via a compactly supported ambient isotopy.

The same strategy can be used to find the homotopy type of the space of embeddings in 2. As Allen Hatcher says in his comment, the space is 2 is not contractible, and one way to explain is that the homeomorphism group of $\mathbb R^2$ that fixes $0$ is not contractible. I think the group is homotopy equivalent to $O(2)$. See the paper above of Yagasaki.