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 $\bar{f}$.
The answer to Q2 is also no. Let $y$ be a nonprincipal ultrafilter on $\mathbb{N}$. The sets $A \times A\setminus\Delta$ where $A \in y$ and $\Delta = \{(n,n) : n \in \mathbb{N}\}$ form a filter base on $\mathbb{N}\times\mathbb{N}$. Let $x$ be an ultrafilter on $\mathbb{N}\times\mathbb{N}$ that contains all these sets. The left and right projections $\pi_1,\pi_2:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ both send $x$ to $y$, but they are not equal on any neighborhood of $x$.
However, the answer to Q2 is yes when $x$ is a selective ultrafilter. Recall that $x$ is selective if for every $h:\mathbb{N}\to\mathbb{N}$ there is a set $A \in x$ on which $h$ is either constant or one-to-one. Given $f,g:\mathbb{N}\to\mathbb{N}$ such that $\bar{f}(x) = \bar{g}(x)$ is nonprincipal, then we can find $A \in x$ on which $f$ and $g$ are both one-to-one. In that case, $f\circ g^{-1}$ must be well-defined on some $A' \in x$. Any extension of $f\circ g^{-1}$ to the complement of $A'$ must map $x$ to $x$, which means that $f \circ g^{-1}$ is the identity on some $A'' \in x$. Thus $f$ and $g$ are equal on the neighborhood of $x$ defined by $A''$.