Suppose that $(M,\omega)$ is a a symplectic manifold. Since $\omega$ is non-degenerate, it sets up an isomorphism $$\omega:TM\rightarrow T^*M$$ between $TM$ and $T^*M$. Why does non-degeneracy ($\omega\wedge\omega\wedge\cdots\wedge\omega\neq0$) imply this isomorphism?
Also, does $\mathcal L_{X_H}\omega=0$ alone enough to imply that we are dealing with a Hamiltonian VF?
On
The non degeneracy of $\omega$ gives the injectivity of this map $$T M\to T^\ast M \ , X\mapsto \omega(X,\cdot)$$ Since for each $p\in M, T_pM\cong T_p^\ast M$ any injective function from $T_p M$ to $T_p^\ast M$ is bijective.
For your second question your notation is not quite correct. You shouldn't call it $X_H$ because that implies the vector field is Hamiltonian with Hamiltonian function $H$. I agree with Romero's answer about the cohomology group. However, you can conclude by Poincare's lemma (closed implies locally exact) that if $\mathcal L_X\omega=0$ then for every $p\in M$ there exists a neighbourhood $U\subset M$ containing $p$ and $f\in C^\infty(U)$ such that on $U$ we have $X=X_f$.
You need to compute the determinant of the linear map associated to $\omega$ and prove that it is nonzero, right? First, try to use what you know about $\omega\wedge\cdots\wedge\omega$ at a point $p\in M$ after contracting it with a basis $\{X_1,\dots,X_{2n}\}\subset T_pM$.
About the Lie derivative, the condition $\mathcal{L}_X\omega=0$ only ensures that $\mathrm{d}(\imath_X\omega)=0$, and if the first cohomology group of $M$ is nontrivial, one cannot guarantee that it is a hamiltonian vector field.
A example of this is the vector field generating the action of $S^1$ over $S^1\times S^1$: take the area form for the torus $S^1\times S^1$ and define the action by $(z,x,y)\mapsto (z\cdot x,y)$, with $x,y,z\in S^1$. This vector field satisfies $\mathcal{L}_X\omega=0$, but there is no function $f$ such that $\imath_X\omega=-\mathrm{d}f$.