There are two things going on here. The first is that you are taking the tensor product of two vector spaces. The second is that you have Lie algebra structures.
For any two vector spaces, we have an isomorphism $V\otimes W \to W\otimes V$ given by $v\otimes w \mapsto w\otimes v$. We can easily get that this exists and is an isomorphism by using the universal property: a bilinear map from $V\times W$ is easily turned into a binlinear map from $W\times V$ by defining $\widetilde{f}(w,v)=f(v,w)$. If you have not seen this before, you should try to work out the details.
The issue that remains is whether this isomorphism of vector spaces is compatible with the Lie algebra structure. Let $\tau(v\otimes w)=w\otimes v$. Then we must show that for each $g\in \mathfrak g$, $\tau(g.v\otimes w)=g.\tau(v\otimes w)$. Note that by linearity, it suffices to check the statement on simple tensors. This is a straightforward calculation, using only the definition of the $\mathfrak g$-module structure on the tensor product.
More generally, what is going on is that the universal enveloping algebra $\mathcal U\mathfrak g$ is a Hopf algebra with comultiplication $\Delta:\mathcal U\mathfrak g\to \mathcal U\mathfrak g\otimes \mathcal U\mathfrak g$ defined on generators by the formula $\Delta(g)=g\otimes 1 + 1 \otimes g$ and extended by the condition that $\Delta$ is a map of algebras (exercise: verify that this map is well defined).
The Hopf algebra structure is what allows us to define an action on $V\otimes W$. Because a $\mathfrak g$-module is the same thing as a $\mathcal U\mathfrak g$-module, $V\otimes W$ is a priori only a module over $\mathcal U\mathfrak g\otimes \mathcal U\mathfrak g$. To get the structure of a $\mathcal U\mathfrak g$-module, we must pull back along $\Delta$. The module structure comes from the composite
$$\mathcal U\mathfrak g\to \mathcal U\mathfrak g\otimes \mathcal U\mathfrak g \to \operatorname{End}(V)\otimes \operatorname{End}(W) \to \operatorname{End}(V\otimes W).$$
where all the maps are maps of algebras. The fact that $\tau$ is a map of $\mathfrak g$-modules comes from the easy to verify fact that $\tau \Delta = \Delta \tau$ (where $\tau$ in this context is the swapping map on $\mathcal U\mathfrak g \otimes \mathcal U\mathfrak g$).
In general, the category of modules over a cocommutative Hopf algebra, or comodules over a commutative Hopf algebra will have a commutative tensor product induced from the tensor product of vector spaces. More generally, one can study quantum groups, where the commutativity or cocommutativity is weakened so that instead of the category of representations being symmetric monoidal, it is braided monoidal. In this way, understanding quantum groups allows one to better understand representations of the braid group.
Just to give a definite answer :
Q1 : No. For example, any simple modules can be expressed as some quotient of $U(\mathfrak g)$ (look for "Verma modules"), and there is so such module in $Rep(\mathfrak g)$.
Q2 : No. For example, the natural sequence $$ 0 \to M(-2) \to M(0) \to L(0) \to 0 $$ does not split (here $M(-2), M(0)$ are two Verma modules for $\mathfrak{sl}_2$ and $L(0)$ is the trivial representation). Note that $M(0), M(-2)$ are infinite-dimensional.
Q3 : The only relation is that $Rep(\mathfrak g)$ is a semisimple category when $\mathfrak g$ is semisimple. So your last sentence is the correct interpretation.
Remark : $U(\mathfrak g)$ is a rather complicated ring. For more information you can look at Dixmier's book on Universal Envelopping algebra. Also you can look at Humphrey's book about BGG category $\mathcal O$.
Best Answer
This is not such a trivial problem. The solution I know uses the Jacobson density theorem: Let $M$ be a simple $R$-module, $D = \mathrm{End}_R(M)$. If $\varphi \in \mathrm{End}_D(M)$, $x_1,\ldots, x_n \in M$, then there exists $r \in R$ such that $rx_i = \varphi(x_i)$ for all $i$. I will show how this reduces to the case of principal tensors.
Let $\mathcal U_1$ and $\mathcal U_2$ be the enveloping algebras of $\mathfrak g_1$ and $\mathfrak g_2$. By Dixmier's Lemma, even if $V_1$ and $V_2$ are infinite-dimensional, $$ \mathrm{End}_{U_i}(V_i) = \mathbb C$$ for $i = 1,2.$ (This is a generalization of the usual finite-dimensional Schur lemma to modules of countable dimension over $\mathbb C$). Now suppose that $$v = \sum_{i=1}^n v^1_i \otimes v^2_i \in V_1 \otimes V_2$$ is nonzero. We want to show $\mathcal U(\mathfrak g_1 \oplus \mathfrak g_2) v = V_1 \otimes V_2$. Without loss of generality, we may choose $\{v^1_i\}$ to be linearly independent over $\mathbb C$ and $v^2_i \neq 0$ for all $i$. By Jacobson density, there exists $u \in \mathcal U_1$ such that $$ uv^1_i = \begin{cases} v^1_1 & i= 1 \\ 0 & i > 1 .\end{cases}$$ Hence for $u\otimes 1 \in \mathcal U_1 \otimes \mathcal U_2 =\mathcal U(\mathfrak g_1 \oplus \mathfrak g_2)$, $$(u \otimes 1)v = v^1_1 \otimes v^2_1,$$ which reduces to the case of when $v$ is a principal tensor.