Why can we define the moment map in this way (i.e. why is this form exact)? Given a symplectic manifold $(X, \omega)$ and a group $G$ acting on $X$ preserving the symplectic form, we define the moment map $\mu : X \to \mathfrak{g}^*$ so that
$$
\langle d\mu(v), \xi\rangle = \omega\big(\xi^*(x), v\big)
$$
where $\xi \in \mathfrak{g}$, $\xi^*$ is the vector field generated by $\xi$, and $v \in T_x(X)$.
Why can we do this? Why is $\mu$ defined?
I read the above definition as follows. Define the $\mathfrak{g}^*$-valued 1-form $\theta$ via
$$
\theta(v) = \omega(\xi^*,v).
$$
Then this form is exact, and so we define $\mu$ so that $d\mu = \theta$.
Why can we do this? Having done some simple computations, the fact that $G$ preserves the symplectic form seems to be imply that exactness holds (and from the above definition it is necessary), but I don't know why this should be the case.
My question is:

Why does the fact that $G$ preserves the symplectic form imply that the 1-form $\theta$ defined above is exact?

Also:

Is the preservation of the symplectic form equivalent to the exactness of $\theta$?

 A: One thing that seems to be confusing you is that there is no "the moment map."  There often several different moment maps for the same action, and as Jose says, sometimes none.
On the other hand, your questions seem to really be 

If X is a vector field, are the conditions
  
  
*
  
*$X$ preserves $\omega$ and
  
*$i_X\omega=\omega(X,-)$ is an exact 1-form
  
  
  equivalent?" 

The answer is "no", but it's "yes" if you replace "exact" by "closed".  The preservation of $\omega$ by $X$ is the same as saying that the Lie derivative $\mathcal{L}_X\omega=0$.  By Cartan's magic formula,
$$\mathcal{L}_X\omega=d(i_X\omega)+i_X(d\omega)=d(i_X\omega)$$ since the second term is zero by the closedness of $\omega$.
This distinction is important, since lots of symplectic actions don't have moment maps because of the existence of $H^1$.  Think about, for example, the 2-torus $T^2$ acting on itself; any invariant volume form can also be thought of as an invariant symplectic form, bu the map $\mathfrak t^2\to H^1(T^2)$ given by $X\mapsto [i_X\omega]$ is an isomorphism.  No non-trivial element of the Lie algebra acts by a Hamiltonian vector field, even though they all act by symplectic ones. 
A: There is yet another interpretation of the already mentioned obstructions for the existence and uniqueness of momentum maps (moment mappings, moment maps, momentum mappings....) in terms of Poisson geometry. This even generalizes then to Poisson manifolds and makes things perhaps a little bit more transparent:
For the existence of a (non-equivariant) momentum map you ask Poisson vector fields to be Hamiltonian, the obstruction therefore lies in the first Poisson cohomology, which is precisely the quotient of Poisson vector field modulo Hamiltonian vector fields.
In the symplectic case this quotient becomes canonically isomorphic to the first deRham cohomology as symplecticl vector fields correspond to closed one-forms while Hamiltonian ones correspond to exact one-forms via the musical isomorphism induced by $\omega$.
The uniqueness (and also the equivariance) is then controlled by the zeroth Poisson cohomology which are th functions with trivial Hamiltonian vector field, also called the Poisson center. In the symplectic and connected case, these are just the constant functions, but in the general Poisson case this might be a much more interesting cohomology. 
The advantage of this point of view is, perhaps, the fact that now all obstructions are somehow arising from the same complex: the canonical complex associated to a Poisson structure $\pi$. As a vector space this is just the sections of the exterior algebra of the tangent bundle (i.e. the multivector fields) and the differential is the Schouten bracet with $\pi$. In the symplectic case, this boils down to the deRham complex via the musical isomorphim.
You can find a detailed discussion of this in the (by the way very: nice) book of Ana Cannas da Silva and Alan Weinstein on "Geometric Models for Noncommutative Algebras".
A: Both answers are "No."
There are well-known obstructions to the existence of an equivariant momentum mapping arising from the action by symplectomorphisms of a group $G$ on a symplectic manifold.  They can be phrased in many ways, but if $G$ is connected and its Lie algebra is semisimple, for example, the obstructions vanish.
A nice treatment can be found in the classic paper by Atiyah and Bott: "The moment map and equivariant cohomology" in Topology 1984.
After-dinner update:
Let me try to add some more details, since it seems I was a little too quick both reading the question and also the comments! (In my defense, I was getting really hungry and wanted to get home!)
Let $(M,\omega)$ be a symplectic manifold and let $G$ be a connected Lie group acting on $M$ via symplectomorphisms.  Let $\mathfrak{g}$ be the Lie algebra of $G$ and for every $X \in \mathfrak{g}$ let $\xi_X$ denote the corresponding vector field on $M$.  Since $G$ acts symplectomorphically, $\xi_X$ is a symplectic vector field; that is, $\mathcal{L}_{\xi_X} \omega = 0$, which, as Ben points out, implies that $d i_{\xi_X}\omega = 0$.  Let $\mathrm{Sym}(M)$ denote the Lie algebra of symplectic vector fields on $M$.  We have a Lie algebra homomorphism $\mathfrak{g} \to \mathrm{Sym}(M)$.
A symplectic vector field $\xi$ is said to be hamiltonian if $i_\xi\omega$ is not merely closed, but also exact.  The $G$-action is hamiltonian if the $\xi_X$ are hamiltonian for all $X \in \mathfrak{g}$.  Let $\mathrm{Ham}(M)$ denote the Lie algebra of hamiltonian vector fields.  It is not hard to show that the Lie bracket of two symplectic vector fields is hamiltonian, so $\mathrm{Ham}(M)$ is an ideal in $\mathrm{Sym}(M)$ with abelian quotient.  This gives rise to a short exact sequence of Lie algebras
$$
0 \longrightarrow \mathrm{Ham}(M) \longrightarrow \mathrm{Sym}(M) \longrightarrow H^1(M) \longrightarrow 0
$$
where $H^1(M)$ is the first de Rham cohomology group thought of as an abelian Lie algebra.
If $\xi \in \mathrm{Ham}(M)$, then $i_\xi \omega = df$ for some $f \in C^\infty(M)$.  This defines a map $C^\infty(M) \to \mathrm{Ham}(M)$ which is also a Lie algebra homomorphism if we make $C^\infty(M)$ into a Lie algebra via the Poisson bracket.  The kernel of this map consists of the locally constant functions, whence we have another short exact sequence of Lie algebras
$$
0 \longrightarrow H^0(M) \longrightarrow C^\infty(M) \longrightarrow \mathrm{Ham}(M) \longrightarrow 0
$$
Putting these two sequences together we have a four-term exact sequence starting and ending at the first two de Rham cohomology groups:
$$
0 \longrightarrow H^0(M) \longrightarrow C^\infty(M) \longrightarrow \mathrm{Sym}(M) \longrightarrow H^1(M) \longrightarrow 0
$$
Now the symplectic $G$-action defines a Lie algebra homomorphism $\mathfrak{g} \to \mathrm{Sym}(M)$ and for the existence of a momentum map, we want this to lift to a Lie algebra morphism $\mathfrak{g} \to C^\infty(M)$.
There is an immediate obstruction for the map to lift at the level of vector spaces, namely the map $\mathfrak{g} \to \mathrm{Sym}(M) \to H^1(M)$ is a Lie algebra cocycle with values in the trivial module $H^1(M)$, and so defines a class in $H^1(\mathfrak{g};H^1(M))$.  If this class vanishes, we do get a map $\mathfrak{g} \to C^\infty(M)$ which may only be a homomorphism modulo $H^0(M)$.  In other words, it defines a Lie algebra homomorphism to a central extension of $C^\infty(M)$ defined by a 2-cocycle with values in the trivial module $H^0(M)$.  The moment map will exist if the class of this cocycle in $H^2(\mathfrak{g}; H^0(M))$ vanishes.
For example, if $\mathfrak{g}$ is semisimple, then both $H^1$ and $H^2$ vanish and the momentum map exists.
The beautiful observation of Atiyah and Bott is that the obstruction can be reinterpreted in terms of the Cartan model for the equivariant de Rham cohomology of $M$, where it becomes simply the obstruction to extending $\omega$ to an equivariant cocycle.
