If $M$ has hyper-kaehler structure then $M//G$ has hyper-kaehler structure? 
A) Let $M$ is a non-compact manifold and $G$ be a compact Lie group which
  acts on $M$ and preserves complex structure then If $M$ has Kaehler
  manifold, then the symplectic quotient of $M$, i.e, $M//G$ has kaehler
  structure?

What about hyper-kaehler? 

B) If $M$ has hyper-kaehler structure then $M//G$ has hyper-kaehler
  structure?

 A: If $(X,\omega)=(M,I,\omega)$ is Kaehler, and $G$ is a compact Lie group, which acts freely on $M$ and preserves both $I$ and $\omega$, then the symplectic quotient is Kaehler. More precisely, $M//G\simeq X^{st}/G^c$, where $G^c$ is the complexification of $G$, and $X^{st}\subset X$ are the stable points for the action. See, e.g., HKLR, Theorem 3.1 
and Hitchin's Sem. Bourbaki 748, section 3.4.
In the hyperkaehler case, there is again compatibility with the symplectic quotient, but it is more subtle. Your suggestion in B) cannot possibly work, because
$\dim_R M-2\dim_R G$ need not be divisible by $4$. The actual relation is as follows. Suppose $M$ is hyperkaehler, and $G$ (acts freely in a hamiltonian way by isometries and) preserves "the" three Kaehler forms $\omega_I$, $\omega_J$, $\omega_K$. Then
we have a moment map
$$
\mu=(\mu_I,\mu_J,\mu_K): M\to \mathbb{R}^3\otimes\mathfrak{g}^\vee,
$$
and for a regular value $\xi$, the quotient $\mu^{-1}(\xi)/G$ is hyperkaehler (see HKLR, Theorem 3.2 and SB 748, Theorem 3.) added: provided $\xi$ is $G$-invariant. 
If you want to link this with the symplectic or Kaehler quotient, you  fix one of the complex structures, say $I$, and break the three moment maps into a real moment map $\mu_I$ and a complex one, $\mu_c=\mu_J+i\mu_K$. Then, if
$\xi=(\xi_1,\xi_2,\xi_3)$, we have
$$
\mu^{-1}(\xi) = \mu_I^{-1}(\xi_1)\bigcap\mu_c^{-1}(z)
$$
where $z=\xi_2+i\xi_3$.
The hyperkaehler quotient is the symplectic reduction of $\mu_c^{-1}(z)$,  or equivalently, the Kaehler quotient
$\mu_c^{-1}(z)^{st}/G^c$.
A: No, I think this need not be the case. Consider the usual action of $S^1$ on $\mathbb{C}^2$. The symplectic quotient is $\mathbb{P}^1$, which is not hyper-Kahler for dimension reasons.
A: There is a notion called hyperkahler quotient, if it is what you ask. (cf. Hitchin, etc., Hyperkahler Metrics and Supersymmetry, Commun. Math. Phys. 108, 535-589 (1987)).  Assume that G is a compact Lie group acting freely on a hyperkahler manifold M, which preserves the complex structures I, J, K, and corresponding Kahler forms $\omega_{I}$,   $\omega_{J}$ and $\omega_{K}$.  If we have three moment maps $h_{I}$ $h_{J}$  $h_{K}$, then we denote $$h=(h_{I},h_{J},h_{K}): M \rightarrow g^{*} \otimes R^{3},$$ and $h^{-1}(0)/G$ is a hyperkahlerian. See Theorem 3.2 in the above paper. 
