Finding a quotient of the free group Consider the free group $F_S$ over a set $S$. Let $x \neq e$ be an element of $F_S$. Is it true that there is a group morphism $\varphi : F_S \to G$ to a finite group $G$ such that $\varphi(x) \neq e$ ?
What happens if, instead of a single element $x \in F_S$, one has finitely many non neutral elements $x_1, \dots, x_n \in F_S$ and asks the $\varphi(x_i)$ to be non neutral ?
 A: Yes, this property is known as residually finiteness.
Let $X$ be a finite set of elements of the free group, all different from the identity. In the Cayley graph of $F_S$, consider the directed subgraph $G$ obtained by reading each element of $X$ (in reduced form) from $1$. For instance, if $u = ab\bar a bba\bar b$, you will get the path 
$$
1 \xrightarrow{a} a \xrightarrow{b} ab \xrightarrow{\bar a} ab\bar a \xrightarrow{b} ab\bar a b \xrightarrow{b} ab\bar a bb \xrightarrow{a} ab\bar a bba \xrightarrow{\bar b} ab\bar a bba\bar b
$$
Now, modify each path by reversing each edge labelled by a letter $\bar a$, where $a \in S$. With the previous example, you would get
$$
1 \xrightarrow{a} a \xrightarrow{b} ab \xleftarrow{a} ab\bar a \xrightarrow{b} ab\bar a b \xrightarrow{b} ab\bar a bb \xrightarrow{a} ab\bar a bba \xleftarrow{b} ab\bar a bba\bar b
$$
Note that if two words of $X$ have a common prefix, their respective paths share a common prefix. Let $Q$ be the (finite) set of vertices of $G$. Then every letter $a$ of $S$ defines a partial function from $Q$ to $Q$ (given by the edges). I claim that this partial function is injective. Indeed if there are
two edges $p \xrightarrow{a} q \xleftarrow{a} r$, then $q = pa$ and $q = r\bar a$, which means that $pa \bar a$ was not a reduced word.
Now, complete each partial injection to a bijection in some arbitrary way.
Then each letter $a \in S$ defines a permutation on $Q$. Let $H$ be the (finite) group generated by these permutations and let $f: F_S \to H$ be the natural morphism. If $x \in X$, then the permutation induced by $x$ maps $1$ onto $x \not= 1$ and thus $f$ separates $1$ from any element of $X$.
A: The Wikipedia article on the Ping Pong Lemma gives a nice proof that the subgroup of $SL(2,\mathbb{Z})$ generated by
$$ A = \begin{pmatrix}1 & 2\\ 0 & 1\end{pmatrix}, \qquad B=\begin{pmatrix}1 & 0\\2 & 1\end{pmatrix}$$
is free of rank two.  By considering maps $\phi_p: SL(2, \mathbb{Z})\rightarrow SL(2, p)$, for increasing primes $p$, it is easy to show $F_2$ is residually finite (the name for the property in your first paragraph).
Since all finitely-generated free groups are subgroups of $F_2$, and this property clearly passes to subgroups [restrict the maps], all finitely-generated free groups are residually finite.
Since any given element only has finitely many generators in its representation, and we can always extend maps trivially to the rest of the generators, all free groups are residually finite.
Finally, if we want to do this for finitely many elements, where we have $\phi_i: F_S\rightarrow G_i$ such that $\phi_i(x_i)\neq 1$, just consider
$$ \prod_i\phi_i: F_S\rightarrow \prod_i G_i$$
