What is the etale fundamental group of Z((x))? I know the fundamental group of $\mathbb{Z}$ is trivial, and the prime candidate for finite etale covers involving the $x$'s is the endomorphism sending $x\mapsto x^n$, but that's ramified at all primes dividing $n$, so my guess is that $\pi_1(\text{Spec }\mathbb{Z}((x)))$ is trivial, though I'm not sure.
What about the fundamental group of things like $\text{Spec }\mathbb{Z}((x))\otimes_\mathbb{Z}R$ where $R$ is some ring like $\mathbb{Z}[1/N]$ or $\mathbb{Z}[1/N,e^{2\pi i/N}]$ or $\mathbb{C}$?
 A: Well, let's do the easiest case: $X=\text{Spec}(\mathbb{C}((x)))$. Since this is a field, we know that $\pi_1(X)=G_{\mathbb{C}((x))}$. But, all extensions of $\mathbb{C}(x))$ are simple:

Claim: All extensions of $\mathbb{C}((x))$ are of the form $\mathbb{C}((t))$ with $t^n=x$ for some $n$.

Proof: Let $K/\mathbb{C}((x))$ be a finite extension, and let $\mathcal{O}_K$ be its ring of integers, with uniformizer $\pi$. Since $\mathbb{C}[[x]]$ has closed residue field, we know that the $f$ of this extension is $1$. This implies that $e=n:=[K:\mathbb{C}((x))]$. Thus, $x=\pi^n u$ for some unit $u\in\mathcal{O}_K^\times$. Now, by Hensel's lemma, the polynomial $T^n-u$ splits into unique linear factors in $\mathcal{O}_K[T]$, and so $u$ has an $n^{\text{th}}$ root $v$ in $\mathcal{O}_K$. Since $\varpi=v\pi$ is another uniformizer, we know that $x=\varpi^n$, where $\pi$ is a uniformizer of $\mathcal{O}_K$. That said, we note that $\mathbb{C}[[x]][\varpi]\subseteq \mathcal{O}_K$ is actually integral, since it's isomorphic to $\mathbb{C}[[T]]$, and thus $\mathcal{O}_K=\mathbb{C}[[x]][\varpi]$. Thus, $K=\mathbb{C}((x))(\varpi)$, which is of the desired form. $\blacksquare$
From this, it's easy to see that 
$$\pi_1(X)=G_{\mathbb{C}((x))}=\varprojlim \text{Gal}\left(\mathbb{C}((\sqrt[n]{x}))/\mathbb{C}((x))\right)=\varprojlim \mathbb{Z}/n\mathbb{Z}=\widehat{\mathbb{Z}}$$
I'll come back later to answer the others.
