A linear represenation of a Lie algebra $\mathfrak{g}$ is a pair $(V, \nabla)$ consisting of a vector space $V$ and a Lie algebra morphism
$$\nabla: \mathfrak{g}\longrightarrow \mathsf{End}(V),$$ where $\mathsf{End}(V)$ is endowed with the bracket $[T_1, T_2]:=T_1\circ T_2-T_2\circ T_1$.
Question 1. Is there a bijection between the set of representations of $\mathfrak{g}$ on $V$ and the set of representations of $\mathfrak{g}$ on $V^*$?
I believe this is the case for given $\nabla$ we can defined $\nabla^*: \mathfrak{g}\longrightarrow \mathsf{End}(V^*)$ setting $$\nabla_x^*(f):=-f\circ \nabla_x.$$ We need this minus sign in order to get a morphism of Lie algebras. On the other hand, given $\nabla^*$ we define $\nabla_x: V\longrightarrow V$ by duality as $$f(\nabla_x v)=-\nabla_x^*(f)(v),\forall f\in V^*.$$
The minus sign here is for the definition to be consistent with $\nabla_x^*$ we defined previously.
I have a second question. I'm guessing any representation $\nabla^*: \mathfrak{g}\longrightarrow \mathsf{End}(V^*)$ induces $$\Lambda^k \nabla^*: \mathfrak{g}\longrightarrow \mathsf{End}(\Lambda^k V^*)$$ by:
$$(\Lambda^k \nabla^*)_x(f_1\wedge \ldots \wedge f_k):=\sum_{j=1}^k f_1\wedge \ldots \wedge \nabla_x^* f_j\wedge\ldots \wedge f_k.$$
This is the only linear extension of $\nabla_x^*$ to $\Lambda^k V^*$. Probabily we can define $\Lambda^k \nabla: \mathfrak{g}\longrightarrow \mathsf{End}(\Lambda^k V)$ by $$(\Lambda^k \nabla)_x(v_1, \ldots, v_k):=-\sum_{j=1}^k v_1\wedge \ldots \wedge \nabla_xv_j\wedge \ldots \wedge v_k,\quad (I)$$ and $\Lambda^k \nabla^*$ will then become the dual representaion of $\Lambda^k \nabla$.
Question 2. Does every representation of $\mathfrak{g}$ on $\Lambda^k V^*$ arises from a representation of $\mathfrak{g}$ on $V^*$ according to $(I)$?
I don't think this is true anyway.
Thanks.