$\newcommand\R{\mathbb R}$For any $(x_1,\dots,x_n)\in\R^n$,
$$P(x):=\prod _{i=1}^{n}(x-x_{i})=\sum _{k=0}^{n}(-1)^{k}e_k(x_1,\dots,x_n)x^{n-k},$$
where the $e_k$'s are the elementary symmetric polynomials in $x_1,\dots,x_n$. Therefore and because "the roots of a polynomial are determined by its coefficients", there is a function $g\colon\R^m\to[\R]^n$ such that
$$[x_1,\dots,x_n]=g(e_0(x_1,\dots,x_n),\dots,e_n(x_1,\dots,x_n))$$
for all $(x_1,\dots,x_n)\in\R^n$, where $m:=n+1$,
$$[x_1,\dots,x_n]:=\{(x_{\pi(1)},\dots,x_{\pi(n)})\colon\pi\in S_n\},$$
$$[\R^n]:=\{[x_1,\dots,x_n]\colon(x_1,\dots,x_n)\in\R^n\},$$
and $S_n$ is the set of all permutations of the set $\{1,\dots,n\}$.
On the other hand, if a function $f\colon\R^n\to\R$ is symmetric, then there clearly exists a function $u\colon[\R^n]\to\R$ such that $$f(x_1,\dots,x_n)=u([x_1,\dots,x_n])$$ for all $(x_1,\dots,x_n)\in\R^n$.
Thus, $$f(x_1,\dots,x_n)=h(e_0(x_1,\dots,x_n),\dots,e_n(x_1,\dots,x_n))$$ for all $(x_1,\dots,x_n)\in\R^n$, where $h:=u\circ g$. This answers your Question 1. $\quad\Box$
As an illustration of the previous consideration, here is the identity $$x\sin y+y\sin x \\ =\Big(-\frac{p(x,y)}2+\sqrt{\frac{p(x,y)^2}4-q(x,y)}\Big) \\ \times\sin\Big(-\frac{p(x,y)}2-\sqrt{\frac{p(x,y)^2}4-q(x,y)}\Big) \\ +\Big(-\frac{p(x,y)}2-\sqrt{\frac{p(x,y)^2}4-q(x,y)}\Big) \\ \times\sin\Big(-\frac{p(x,y)}2+\sqrt{\frac{p(x,y)^2}4-q(x,y)}\Big) \\ $$ for all $(x,y)\in\R^2$, where $p(x,y):=x+y$ and $q(x,y):=xy$ are elementary symmetric polynomials in $x,y$.