Question 1: Let $G$ be a connected reductive group defined over a number field $K$, and $\pi$ is an irreducible cuspidal automorphic representation of $G(\mathbb{A}\_K)$. Then by a theorem of Flath, we have $\pi=\otimes^{'}\pi\_v$. All but finitely many $\pi\_v$ are unramified and they all generic by Shalike.


I want to know whether there exists a finite place $\mu$ such that $\pi\_{\mu}$ is supercuspidal for $G(K\_{\mu})$ or there is no relation with cuspidal automorphic representation for $G(\mathbb{A}\_K)$ and supercuspidal representation of $G(K\_{\mu})$. 


Question 2: Let $S$ be a finite set containing infinite places, and assume $\pi\_v$ is unramified for $v \notin S$ and $\pi\_{\mu}$ is generic for $\mu \in S$, then can I say $\otimes\_{v}^{'}\pi\_v$ is a cuspidal automorphic representation of $G(\mathbb{A}\_K)$? If not, what additional condition should I add?


Question 3: I know there is a proposition below proving by Bernstein-Zelevinsky classification.



> 
> $\pi\_v$ is an unramified and supercuspidal for $GL\_n(K\_v)$, where $v$ is a finite place if and only if $n=1$ and $\pi\_v$ is an unramified quasi-character.
> 
> 
> 


I want to know whether the proposition holds for general connected reductive p-adic group.



Some question about cupidal automorphic representation and supercuspidal representation
It is not necessary that a cuspidal $\pi$ have a supercuspidal local component: for example, let $\Delta(\tau)$ be Ramanujan's delta function and let $\pi = \otimes' \pi\_p$ denote the associated cupsidal automorphic representation of $\textrm{GL}\_2(\mathbb{A}\_{\mathbb{Q}})$ (see Kudla's article [here](https://mathscinet.ams.org/mathscinet/search/publdoc.html?extend=1&l=100&pg1=INDI&s1=213732&sort=Newest&vfpref=html&r=40&mx-pid=1990371)). Since $\Delta$ has level 1, $\pi\_p$ will be unramified for every prime $p$. 


What are your assumptions for question 2? Given a random collection $\{\pi\_v\}\_v$ of irreducible local representations, almost all of which are unramified, the representation $\otimes' \pi\_v$ will be neither cupsidal nor automorphic. 


There are many ways to prove question 3. For simplicity, assume $G$ is unramified over $K\_v$ (that is, quasi-split and split over an unramified extension), let $K$ denote a hyperspecial maximal compact subgroup, $I\subset K$ an Iwahori subgroup, and let $P = MN$ be a minimal parabolic subgroup. If $\pi$ were both ($K$-)unramified and supercuspidal, then Theorem 3.7 in Cartier's article in the Corvallis proceedings would give
$$0\neq \pi^K \subset \pi^I \stackrel{\sim}{\rightarrow}(\pi\_N)^{K\cap M}.$$
If $N\neq 1$, then the right-hand side would be zero by supercuspidality. Therefore $G$ must be a torus.


