Let $\mathcal{H}$ be infinite dimensional Hilbert space and $D(\mathbb{R}^n)$ be the space of smooth complex functions of compact support. Consider the distribution
$T: D(\mathbb{R}^n) \to \mathcal{H}$
such that $T$ has compact support (in distributional sense).
Is it true that $T$ is a distribution of finite order? (The complex-valued distribution of compact support has finite order but I guess this property does not generalize to vector-valued distributions if the considered vector space is infinite dimensional).
Is it true that $T=\partial^\alpha F$ (the derivative in distributional sense) for some continuous function $F$ and multi-index $\alpha$? (Again I suppose that this statement is in general false since the standard proof of this fact for complex-valued distributions uses the fact that $T$ has finite order).
It seems that you mean by vector-valued distribution just a continuous linear $T:\mathscr D(\mathbb R^n) \to H$. If $T$ has compact support $K$ you can extend it to a continuous linear map $\tilde T: C^\infty(\mathbb R^n) \to H$ by setting $\tilde T(f)=T(f\phi)$ where $\phi$ is a test-function which is equal to $1$ in a neighborhood of $K$. But then continuity means $\|\tilde T(f)\|_H \le C \sup\lbrace |\partial^\alpha f(x)|: x\in L, |\alpha|\le m\rbrace$ for some compact set $L$ and some $m\in\mathbb N$. Isn't this the definition of finite order?