$\newcommand\R{\mathbb R}$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][1] 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. $\quad\Box$

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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 $x,y$. 


  [1]: https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial