The answer to (1) is yes. Let us first show
If $I$ is an ideal of $A[x]$ such that $I\cap S=\emptyset$, then $I$ is contained in $\mathfrak m A[x]$ for some maximal ideal $\mathfrak m$ of $A$.
Proof. Note that
$$I\subseteq \sum_{f\in I} c(f)A[x].$$
So if $I$ is not contained in any $\mathfrak mA[x]$, then $A=\sum_{1\le i\le n} c(f_i)$ for some $f_i\in I$. Fix an $m$ big enough and write
$$f_i(x)=a_{i0}+a_{i1}x+...+ a_{im}x^m$$
($a_{im}$ could be zero) and an identity
$$1=\sum_{i\le n, j\le m} \alpha_{ij}a_{ij}, \quad \alpha_{ij}\in A.$$
Consider the polynomial
$$f=\sum_{i,j}\alpha_{ij}f_i(x)x^{m-j} \in I.$$
The term of degree $m$ in $f$ has coefficient equal to $1$. So $f\in S$. Contradiction.
As any $\mathfrak m A[x]$ is prime and has empty intersection with $S$, the above result implies immediately (1).
(2) The map $\phi$ is clearly surjective (consider $\mathfrak qA[x]$ for any $\mathfrak q\in\mathrm{Spec} A$) but is not injective in general. Consider $A=k[t,s]$ over a field $k$. Then $f(x):=t+sx$ generates a prime ideal $\mathfrak p$ of $A[x]$ which doesn't meet $S$ and we have $\mathfrak pS^{-1}A[x]\cap A=\{0\}$. So the generic fiber of $\phi$ has at least two points (in fact infinitely many points).
Note however that $\phi$ is always injective over any maximal ideal $\mathfrak m$ because $\phi^{-1}(\mathfrak m)=\mathfrak mS^{-1}A[x]$.
The argument for the surjectivity of $\phi$ also shows that $\dim S^{-1}A[x]\ge \dim A$. We also have $\dim S^{-1}A[x]\le \dim A[x]=\dim A +1$ when $A$ is noetherian. I don't know whether the equality is possible.
Update: We have $\dim S^{-1}A[x]=\dim A$ when $A$ is noetherian: let $\mathfrak p_0\subset ... \subset \mathfrak p_n$ be a chain of prime ideals of $A[x]$ contained in $A[x]\setminus S$. By (1), we have $\mathfrak p_n\subset \mathfrak mA[x]$ for some maximal ideal $\mathfrak m\subset A$. Then
$$ \mathfrak p_0\subset ... \subset \mathfrak p_n\subset \mathfrak mA[x]+xA[x]$$
is a chain of prime ideals of $A[x]$. Thus $\dim S^{-1}A[x]\le \dim A[x] -1=\dim A$. Your homework is not so easy :).
First, let me explicitly assume $R$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$.
Proposition. The Krull dimension of $R$ is at most $1$.
Proof. Let $\mathfrak{m} = (t)$, and let $\mathfrak{p}$ be prime. We know $\mathfrak{p} \subseteq \mathfrak{m}$, so it is enough to show that either $\mathfrak{p} = \mathfrak{m}$ or $\mathfrak{p} = (0)$. Suppose $\mathfrak{p} \ne \mathfrak{m}$: then $t \notin \mathfrak{p}$. Let $a_0 \in \mathfrak{p}$. For each $a_n$, because $\mathfrak{p}$ is prime, there exists $a_{n+1}$ in $\mathfrak{p}$ such that $a_n = a_{n+1} t$. By the axiom of dependent choice, this yields an infinite ascending sequence of principal ideals:
$$(a_0) \subseteq (a_1) \subseteq (a_2) \subseteq \cdots$$
Since $R$ is noetherian, for $n \gg 0$, we must have $(a_n) = (a_{n+1}) = (a_{n+2}) = \cdots$. Suppose, for a contradiction, that $a_0 \ne 0$. Then, $a_n \ne 0$ and $a_{n+1} \ne 0$, and there is $u \ne 0$ such that $a_{n+1} = a_n u$. But then $a_n = a_{n+1} t = a_n u t$, so cancelling $a_n$ (which we can do because $R$ is an integral domain), we get $1 = u t$, i.e. $t$ is a unit. But then $\mathfrak{m} = R$ – a contradiction. So $a_n = 0$. $\qquad \blacksquare$
Here's an elementary proof which shows why we can reduce to the case where $R$ is an integral domain.
Proposition. Any non-trivial ring $A$ has a minimal prime.
Proof. By Krull's theorem, $A$ has a maximal ideal, which is prime. Let $\Sigma$ be the set of all prime ideals of $A$, partially ordered by inclusion. The intersection of a decereasing chain of prime ideals is a prime ideal, so by Zorn's lemma, $\Sigma$ has a minimal element. $\qquad \blacksquare$
Thus, we can always assume that a maximal chain of prime ideals starts at a minimal prime and ends at a maximal ideal. But if $R$ is a noetherian local ring with principal maximal ideal $\mathfrak{m}$ and $\mathfrak{p}$ is a minimal prime of $R$, then $R / \mathfrak{p}$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$, and $\dim R = \sup_\mathfrak{p} \dim R / \mathfrak{p}$, as $\mathfrak{p}$ varies over the minimal primes.
Update. Georges Elencwajg pointed out in a comment that the first proof actually works without the assumption that $R$ is a domain, because $(1 - u t)$ is always invertible.
Best Answer
In this answer it is mentioned the following useful result: if $x\in\bigcap_{n=1}^{\infty}I^n$, then $x\in xI$, that is, there is $a\in I$ such that $(1-a)x=0$. One uses this result for a prime ideal $P$, and get an element outside of $P$ which multiplied by $x$ gives zero. When this happens for all primes we can conclude that $x=0$.