Existence of orientable finite volume complete cusp hyperbolic 3-manifolds $\mathbb{H}^3 / \Gamma$, where $\Gamma$ has no parabolic generators?

Question

Let $K$ be a hyperbolic knot, i.e., $S^3 - K$ is an orientable finite volume cusp hyperbolic 3 manifold. Let $M=S^3 - K$ then $M= \mathbb{H}^3/\Gamma$, where $\Gamma$ (Kleinian group) is discret subgroup of $PSL(2,\mathbb{C})$ and also $\pi_1(M)$ is isomorphic to $\Gamma$.

Most of the hyperbolic knots I have seen above $\Gamma$ are always generated by parabolic elements, like 4_1 knot, twist knots, etc.. So, natural questions arise

My question is

1. Does there exist a hyperbolic knot such that $\Gamma$ is two generated Kleinian group where the generators are not commutative and not parabolics?
2. More generally does there exist a cusp hyperbolic $3-$ manifold $M=\mathbb{H}^3/\Gamma$ where $\Gamma$ is Kleinian group of two generators, generators of $\Gamma$ are not commute each other and they also not parabolic elements?

My attempted,

I think the answer is yes for 2nd one,

I am taking two loxodromic elements say $\gamma_1$ and $\gamma_2$ such that whose geodesic axis $l_{\gamma_1}$ and $l_{\gamma_2}$ passing from their fixed points far(they don't share the fixed points) from each other, I believe that $\Gamma= <\gamma_1,\gamma_2>$ from a Kleinian group.

Answer 1

Yes, all hyperbolic three-manifold fundamental groups can be generated by loxodromic elements. For, suppose that $\Gamma = \{ \gamma_i \}$ is a generating set. Take $\gamma$, a loxodromic element which is "sufficiently long" compared to the elements $\Gamma$. Then the set $\{ \gamma \} \cup \{\gamma \cdot \gamma_i\}$ generates and consists only of loxodromic elements.

A more delicate argument will give loxodromic generating sets, of size two, of (say) the twist knot groups.