Your statement of the hom-tensor adjunction is incorrect, and we need to use slightly more than the adjunction to prove what you want. In particular, we need that if $V$ is finite dimensional, then $V^**\cong V$ and $\hom(V,W)\cong V^*\otimes W$. Using the fact that $L_{\lambda}$ is finite dimensional, we have $\hom_{\mathfrak g}(\mathbb C, U\otimes L_{\lambda})\cong\hom_{\mathfrak g}(\mathbb C, U\otimes L_{\lambda}^{**}) \cong \hom_{\mathfrak g}(\mathbb C, \hom_{\mathbb C}(L_{\lambda}^*,U))$ Applying the adjunction, this is isomorphic to $\hom_{\mathfrak g}(\mathbb C\otimes L_{\lambda}^*,U)\cong \hom_{\mathfrak g}(L_{\lambda}^*,U)$.
For part (1), we need a few ideas (the upshot of which are given by Jyrki). If $V$ is a representation, I will let $V_{\lambda}$ denote the weight space of weight $\lambda$.
- If $V$ is a finite dimensional representation of $\mathbb g$, $\lambda$ a weight, and $w$ an element of the Weyl group, then $\dim V_{\lambda}=\dim V_{w\lambda}$. We can see this by noting that for each root $\alpha$, we have a copy of $\mathfrak{sl}_2$ which acts on $\bigoplus_i V_{\beta+i\alpha}$ for every weight $\beta$, and standard rep theory for $\mathfrak{sl}_2$ shows that the equality holds when $w=s_{\alpha}$ is a reflection.
- If $V$ is a finite dimensional representation and $\lambda$ is a weight, then $\dim V_{\lambda}=\dim V^*_{-\lambda}$. This can be shown by taking a basis of weight vectors. If $v$ is a weight vector of weight $\lambda$, the corresponding dual vector $v^*$ has weight $-\lambda$.
- Irreducible finite dimensional $\mathfrak g$-reps are classified by integral dominant weights via highest weight vectors.
- If $V$ is irreducible of highest weight $\lambda$, the weights in $V$ will be the convex hull of the weights $w\lambda$, for $w$ in the Weyl group.
Thus, it suffices to show that $L_{\lambda}^*$ has highest weight $-w_0(\lambda)$ where $w_0$ is the longest length word in the Weyl group. As Jyrki said, this follows from the fact that $w_0(\alpha)<0$ for every positive root $\alpha$, as it will be the only dominant weight of the form $-w\lambda$ for $w$ in the Weyl group.