The answer to Q1 is no. This has been well studied in set theory; you're basically asking whether any two non-principal ultrafilters on $\mathbb{N}$ are comparable under the Rudin-Keisler ordering. Variations on your question have led to many, many interesting developments in set theory, but your question Q1 is easy to answer by a cardinality argument.
First note that every $f:\mathbb{N}\to\mathbb{N}$ has a unique extension to a continuous function $\bar{f}:\beta\mathbb{N}\to\beta\mathbb{N}$. Any $x \in \beta\mathbb{N}$ has at most $2^{\aleph_0}$ images through such $\bar{f}$, but there are $2^{2^{\aleph_0}}$ ultrafilters on $\mathbb{N}$, so there are very many $y \in \beta\mathbb{N}$ which are not images of $x$ through such $\bar{f}$.
(I think the answer to Q2 is also no, but I haven't thought about it much.)