Given a tempered distribution $T$, in order to show that it is a Schwartz function, does it suffice to prove for any $f$ Schwartz, $T(f) = \int g f$ for some $g$ Schwartz? Now if $T$ is a tempered distribution such that $(1+|x|^2)^pT(f) \in H^{q}(\mathbb R^n)$ for any $p,q$. Is it true that $T$ is Schwartz?
Directly producing a Schwartz function corresponding to it seems to be difficult since there is no obvious candidate in my opinion. I propose instead it might be easier to show that the usual definition of a Schwartz function holds with respect to distributional derivatives(no idea though whether this suffices). Is that a better way?
Let me clarify some definitions. Let $\mathcal{S}$ denote the set of Schwartz functions on $\mathbb{R}^n$ and $\mathcal{S'}$ denote the set of tempered distributions, that is, continuous linear functionals on $\mathcal{S}$.
$\textbf{Definition}$: We abuse terminology and say that a tempered distribution $T$ is a Schwartz function if there is some $g \in \mathcal{S}$ such that $T(f) = \int fg$ for all $f \in \mathcal{S}$.
So your first question is really just a question about definitions (and the answer is "yes").
Your second question does not make sense. $T(f)$ is a number and not a function. To say what I think you meant, let me make the following definition.
$\textbf{Definition}$: A locally integrable function $g$ is $\textit{tempered}$ if the map $T_g$ given by $f \mapsto \int fg$ for $f \in \mathcal{S}$ is a well-defined, continuous linear functional on $\mathcal{S}$, i.e. if $T_g \in \mathcal{S'}$.
The point is that we can identify some functions with the corresponding tempered distributions. So maybe your second question was supposed to be: if $g$ is tempered and $(1+|x|^2)^p g \in H^q(\mathbb{R}^n)$ for each $p,q$, then is $T_g$ Schwartz, i.e. is $g$ Schwartz? This question of course has nothing to do with distributions, and the answer is "yes". Indeed, $g \in H^q(\mathbb{R}^n)$ implies, by Sobolev embedding, that $g$ is in $C^k$ for each $k$ and thus in $C^\infty$. And it's then Schwartz because of the $(1+|x|^2)^p$ factors (I'll leave this exercise to you).
Hint for exercise: consider $n = 1$ and $g$ non-negative. Then $\frac{\partial}{\partial x} (1+x^2)^p g(x) = p(1+x^2)^{p-1}2xg(x)+(1+x^2)^p g'(x)$, so $(1+x^2)^p g(x) \in L^2$ implies $(1+x^2)^p g'(x) \in L^2$ implies for any $m \ge 1$, $\sup_x (1+x^2)^m g'(x) < \infty$ (by taking $p$ very large based on $g'$ and $m$). Similarly with higher derivatives of $g$.