Concerning your first question: it's really an issue about finite dimensional vector spaces.
(1). If $V$ is a finite dimensional vector space over a field $k$, then one has an evaluation map
$$ e: V^\ast \otimes V \to k $$
defined by $f\otimes v \mapsto f(v)$. Then $$ e \in (V^{\ast} \otimes V)^\ast . $$ There is a canonical isomorphism $V \otimes V^\ast\to (V^\ast \otimes V)^\ast$ given by $$ (v,f) \mapsto ((g,w) \mapsto g(v)f(w)) . $$ Using this isomorphism, we can regard $e$ corresponds to element $e'\in V\otimes V^\ast$. This is the Casimir element.
Remark: $e'$ is characterized by the following property: it is the unique vector in $V\otimes V^{\ast}$ such that for all vector spaces $W$, the map $$ \hom(V,W) \to W\otimes V^\ast $$ given by $f\mapsto (f\otimes 1_{V^\ast})(e')$ is an isomorphism.
(2). More generally, if one is given a perfect pairing: $$ e : U \otimes V \to k $$ (meaning that the adjoint $U \to V^\ast$ is an isomorphism), one gets by the same procedure an element $e'\in V\otimes U$.
(3). $V = \frak g$ is a semi-simple Lie algebra, then the killing form $V\otimes V \to k$ is perfect. So we get by (2) an the (Casimir) element $e' \in V\otimes V$.
