No to the first question. You can make examples where the "fundamental group at infinity" of $A\cup B$ is nontrivial.

Start with a finite complex $X$ that has trivial homology but nontrivial fundamental group. Embed the suspension $\Sigma X$ in $S^d=\mathbb R^d\cup\infty$. The suspension is contractible and is the union of two cones, also contractible. Let $A$ and $B$ be the complements of the two cones. If $d$ is big enough then $A$, $B$, and $A\cap B$ (the complement of $\Sigma X$) can be shown to be diffeomorphic to $\mathbb R^d$ by the $h$-cobordism theorem. But $A\cup B$. the complement of $X$, is such that the complement of a compact set is never simply connected.

No to the first question. You can make examples where the "fundamental group at infinity" of $A\cup B$ is nontrivial.

Start with a finite complex $X$ that has trivial homology but nontrivial fundamental group. Embed the suspension $\Sigma X$ in $S^d=\mathbb R^d\cup\infty$. The suspension is contractible and is the union of two cones, also contractible. Let $A$ and $B$ be the complements of the two cones. If $d$ is big enough then $A$, $B$, and $A\cap B$ (the complement of $\Sigma X$) can be shown to be diffeomorphic to $\mathbb R^d$ by the $h$-cobordism theorem. But $A\cup B$, the complement of $X$, is such that the complement of a compact set in it is never simply connected.

EDIT:

To be a bit more precise, the open set $A$ (or $B$ or $A\cap B$) will be simply connected if the codimension of its complement in $S^d$ is at least $3$. (This is clear at least if the complement is embedded nicely enough.) And $A$ will have trivial homology, by Alexander duality, so it will be contractible. Now, Siebenmann's thesis gives sufficient conditions on a noncompact smooth manifold (of not too small dimension) for there to be a compact manifold with boundary having that as its interior. If there are arbitrarily small simply connected neighborhoods of infinity then these conditions are satisfied. Using this you see that $A$ is the interior of a contractible manifold with simply connected boundary. The $h$-cobordism theorem implies that that compact thing is a closed disk, so that $A$ is an open disk.

Or you can take a slightly different approach and avoid Siebenmann's thesis. Embed that suspension of $X$ piecewise linearly, make a regular neighborhood that is the union of regular neighborhoods of the two cones intersecting in a regular neighborhood of $X$. Use the complements of these compact neighborhoods.

Or, better yet, let $Q$ be a compact codimension zero acyclic non simply connected manifold (with boundary) in the interior of $D^d$. The complement in $S^{d+1}=(\mathbb R^d\times \mathbb R)\cup\infty$ of $D^d\times [0,1]\cup Q\times [1,2]$ is an open ball, as is the complement of $Q\times [1,2]\cup D^d\times [2,3]$, as is the complement of their union. But the complement of the intersection is not.