When a G-Manifold is a Hamiltonian G-manifold Let G be a lie group, then When a G-Manifold is a Hamiltonian G-manifold and under which condition a manifold is Hamiltonian G manifold for some lie group G
 A: This is a hard question and the answer is not known in general, even for $G = S^1$. The obvious necessary condition for a $G$-action on a compact symplectic manifold $(M, \omega)$ to be Hamiltonian is that the action has fixed points. This is because the Hamiltonian function will have critical points on a compact manifold, and these critical points will correspond to fixed points of the corresponding Hamiltonian action.
A well-known theorem of Ono shows that the existence of fixed points is also a sufficient condition in the case of circle actions on symplectic manifolds of "Lefschetz type."

Theorem. (Ono) Let $(M^{2n}, \omega)$ be a compact symplectic manifold such that the map
  $$\wedge \omega^{n-1}: H^1(M; \Bbb R) \longrightarrow H^{2n-1}(M; \Bbb R)$$
  is an isomorphism. Then a symplectic $S^1$-action on $(M, \omega)$ is Hamiltonian if and only if it has fixed points.

All Kähler manifolds satisfy the hypothesis of the above theorem, so any symplectic $S^1$-action on a Kähler manifold with fixed points is Hamiltonian. However, McDuff found an example of a $6$-dimensional symplectic manifold (not of Lefschetz type) with a symplectic $S^1$-action possessing a fixed point that is not Hamiltonian. So Ono's theorem has no hope of extending to all symplectic manifolds.
An often studied class of symplectic manifolds are the monotone symplectic manifolds. The situation is better for them:

Theorem. Any symplectic $S^1$-action on a monotone symplectic manifold is Hamiltonian.

In the noncompact case, there's an easy general result for exact symplectic manifolds $(M, \omega)$, i.e. where $\omega = -d\theta$ for some $1$-form $\theta$. A symplectic $G$-action on such an exact symplectic manifold is called exact if the action preserves the $1$-form $\theta$.

Theorem. Any exact symplectic action of a compact Lie group $G$ on an exact symplectic manifold $(M, -d\theta)$ is Hamiltonian with Hamiltonian function $H_\xi = \iota(X_\xi)\theta$ for $\xi \in \mathfrak{g}$.

A: It might be useful for interested readers to leave more theorems related to this question. They are all standard results and easy to find in textbooks, and this is why I shall just state them without proofs or references.
To fix notation (there are various conventions in the literature): a symplectic action of $G$ on the symplectic manifold $(M,\omega)$ is said to be hamiltonian if the image of the Lie algebra of $G$ by the induced infinitesimal action lies inside the Lie algebra of hamiltonian vector fields of $(M,\omega)$. A moment map for a hamiltonian action is said to be equivariant if it induces a Lie algebra antihomomorphism between the Lie algebra of $G$ and the poisson algebra of functions of $(M,\omega)$. 
0- If the first cohomology group of the symplectic manifold is trivial, then any symplectic action is hamiltonian.
1- Any hamiltonian action admits a moment map.
2- If the first Chevalley cohomology group of the Lie algebra of $G$ is trivial, then any symplectic action of $G$ is hamiltonian. If, in addition, the second Chevalley cohomology group of the Lie algebra of $G$ is trivial, then any symplectic action of $G$ is hamiltonian and admits an equivariant moment map.
3- Any symplectic action (the Lie group can be noncompact) on an exact symplectic manifold $(M,\omega=\mathrm{d}\theta)$ preserving $\theta$ is hamiltonian and admits an equivariant moment map.   
4- (Gotay and Tuynman) A hamiltonian action of a compact Lie group admits an equivariant moment map (the manifold does not need to be compact).
5- A hamiltonian action on a compact symplectic manifold admits an equivariant moment map.
I am assuming that the manifolds and Lie groups are smooth, Hausdorff, paracompact and connected. 
