1.Intuitively, no. But I can't think of examples.
2.In the Topology Book by GF Simmons, before introducing to stone Weierstrass theorem it is stated, that it is meaningless to speak of polynomials on $X$. And hence they have been introduced as subalgebra generated by $\{1,x\}$. Why can't we just say that they are vector spaces formed by basis $\{1,x,x^2,\ldots\}$?
Here's a general result which answers your first question:
Fact: Suppose $f:X\rightarrow X$. Then the following are equivalent:
$f$ is either the identity function or is constant.
For every topology $\tau$ on $X$, $f$ as a map from $(X,\tau)$ to $(X,\tau)$ is continuous. (Note that we use the the same topology on the domain and the codomain; this is crucial, since e.g. if $X$ has at least two elements then the identity map on $X$ is not continuous with respect to the indiscrete topology on the domain and the discrete topology on the codomain.)
Proof: Clearly if $f$ is either the identity or constant then $f$ is continuous with respect to every topology on $X$. Conversely, suppose there are $x, y\in X$ such that $f(x)\not=x$ and $f(y)\not=f(x)$ (note that this happens exactly when $f$ is neither the identity nor constant). Consider the topology $$\tau=\{X,\emptyset,\{f(x)\}\}.$$ The set $A=\{f^{-1}(\{f(x)\})\}$ contains $x$ (so $A\not=\emptyset$) and $x\not=f(x)$ (so $A\not=\{f(x)\}$) and doesn't contain $y$ (so $A\not=X$); this means $A$ isn't open in the $\tau$ sense. But $A$ is the $f$-preimage of a set which is open in the $\tau$ sense, so $f$ is not continuous with respect to $\tau$. $\quad\quad\Box$