Exact Group Representation Definition: Exact Representation of finite group $G$ in some field $V$ - is injective homomorphism
$$\rho : G\to GL(V).$$
(I don't know English terminology, so you may correct me; or probably exists other definition of "exact representation"(do it exist?)).
Question: Do exists non-exact (in my terminology) representations?
 A: Let me summarize the comments (leaving out the snap-discussion):
An injective representation is usually called faithful while geometers often prefer to use effective (but more in connection with group actions, as far as I know).
It is easy to build non-faithful representations:
Take a normal subgroup $N \neq \{0\}$ and look at any representation $\rho: G/N \to \operatorname{GL}{(V)}$. Then the composition $\rho \circ \pi$ of $\rho$ with the canonical projection $\pi: G \to G/N$ will have $N \subset \ker{(\rho \circ \pi)}$. Every representation of $G$ is of this form by the homomorphism theorem. Note also that this shows that representation of a simple group $G$ is either trivial or faithful.
You asked in the comments how one can show that every (finite) group admits a faithful representation. You suggested to use the Cayley theorem, and that's essentially the way to go. You can (sort of) avoid it by taking the vector space $V = k[G]$ with a basis $\{e_{h}\}_{h \in G}$ and let $\lambda: G \to \operatorname{GL}{(V)}$ be the (left) regular representation $\lambda(g)e_h = e_{gh}$. Of course, this is only half avoiding Cayley's theorem, as the proof of Cayley's theorem also relies on the left action of $G$ on itself.
A: In fact, there's a very important class of representations which are typically not faithful (non-exact, in your terminology): characters.
A character of a group $G$ is a homomorphism
$$
\chi:G\longrightarrow{\Bbb C}^\times.
$$
It corresponds to the case where $\dim V=1$.
The set $\hat G$ of caracters of $G$ is itself a group: you can multiply characters by $(\chi_1\cdot\chi_2)(g)=\chi_1(g)\chi_2(g)$
Some groups have very few characters. For instance, the symmetric group ${\cal S}_n$ has only two characters, the constant map and the sign map
$$
{\rm sgn}:{\cal S}_n\longrightarrow\{\pm1\}
$$
giving the sign of a permutation. The alternating group $A_n$ with $n\geq5$ has no non-trivial characters, as does every group lacking non-trivial normal subgroups.
On the other hand, if $G$ is abelian there are usually lots of characters. For instance, it s  not too hard exercise to prove that if $G$ is finite, then ${\hat G}\simeq G$ (although non canonically).
