Is the tensor product (of vector spaces) commutative?

No, it is not commutative. It would imply that all bilinear maps are symmetric.

For any vector space $V$ over a field $K$, we only have an isomorphism \begin{align}V\otimes _KV&\longrightarrow V\otimes_KV, \\v_1\otimes v_2&\longmapsto v_2\otimes v_1. \end{align}

Furthermore, the quotient of $V\otimes_K V$ by the subspace generated by all tensors $v_1\otimes v_2 - v_2\otimes v_1$ is called the symmetric product of $V$ by itself.

Certainly not. We can dualize an element of $\Bbb R^n \otimes \Bbb R^n$ and then view it as a map $(\Bbb R^n)^* \times (\Bbb R^n)^* \to \Bbb R$. Then, if $(\epsilon^a)$ denotes the basis of $(\Bbb R^n)^*$ dual to $(e_a)$, concretely we have $$(e_1 \otimes e_2)(\epsilon^1, \epsilon^2) = e_1(\epsilon^1) e_2(\epsilon^2) = (1) (1) = 1$$ but $$(e_2 \otimes e_1)(\epsilon^1, \epsilon^2) = e_2(\epsilon^1) e_1(\epsilon^2) = (0) (0) = 0 .$$ Therefore $$e_1 \otimes e_2 \neq e_2 \otimes e_1 .$$