I think both can be proved without choice, essentially because, in both cases, whenever you're tempted to choose a basis, you can manage with a little care to get by with a basis of a finite dimensional subspace.
For (2), if there's a linear dependence between the $v_i\otimes w_j$ then it involves only finitely many $v_i$ and $w_j$. Also, the linear dependence must be a (finite) linear combination of the usual relations such as $(u+u')\otimes v-u\otimes v-u'\otimes v$ for the tensor product, so there are finite dimensional subspaces $V'\leq V$ and $W'\leq W$ so that you have the same linear dependence in $V'\otimes W'$. And now you can use bases without invoking choice.
For (1), an element of the kernel is a finite sum of simple tensors $\varphi\otimes\psi$. By choosing a basis of the finite-dimensional subspaces of $V^*$ and $W^*$ spanned by the $\varphi$ and $\psi$ that occur, we can write the element of the kernel as a linear combination of $\{\varphi_i\otimes\psi_j\}_{i,j}$, where $\{\varphi_i\}_i$ and $\{\psi_j\}_j$ are finite linearly independent subsets of $V^*$ and $W^*$.
Now, again without choice, we can find dimensional subspaces $V'\leq V$ and $W'\leq W$ together with bases $\{v_i\}_i$ and $\{w_j\}_j$ that are dual bases to the restrictions of $\{\varphi\}_i$ and $\{\psi\}_j$, and prove that the kernel element is zero using these bases.