No, this is not true. Schwarz distributions with compact support inovlve only finitely many differentiations (every such distribiution is a derivative of some order of a usual, integrable function). Hyperfunctions may involve infinitely many differentiations. For example, in dimension $1$ you can take any entire function $f(x)=\sum a_nx^n$, and then $F=\sum a_n\delta^{(n)}$ is a hyperfunction with support at $0$.

No, this is not true. Schwartz distributions with compact support inovlve only finitely many differentiations (every such distribiution is a derivative of some order of a usual, integrable function). Hyperfunctions may involve infinitely many differentiations. For example, in dimension $1$ you can take any entire function $f(x)=\sum a_nx^n$, and then $F=\sum a_n\delta^{(n)}$ is a hyperfunction with support at $0$.