Fix a global field $F$. Does every automorphic representation of $GL(n)$ appear as an arbitrary twists in the continuous spectrum of $GL(m)$, $m>n$?
What happens for the automorphic representations of $G$ reductive, do they appear in $GL(n)$ for $n$ big?
In general, I don't think that a representation of $GL(n)$ can be viewed directly as a representation of $GL(m)$ (for $m > n$), so it doesn't quite make sense to say that it appears in the continuous spectrum of $GL(m)$. What happens is that automorphic representations on $GL(n_i)$ (where $\sum_i n_i = m$) can be combined into a representations of $GL(m)$ via parabolic induction, and these will appear in appear in the automorphic spectrum of $GL(m)$. (See Langlands's article On the notion of an automorphic representation.)
Some, but not all, of these parabolically induced representations will be in the continuous part of $L^2$ for $GL(m)$. Roughly, the idea is that one can twist the representations of the various $GL(n_i)$-representations independently, and so the parabolically induced representations form a family parameterized by these twists. A certain subfamily will lie in the continuous spectrum for L^2$.
E.g. if we were inducing a pair of characters on $GL(1)$ to get reps. of $GL(2)$, the (suitably normalized) inductions of the unitary characters would lie in the continuous spectrum of $L^2$.