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 finite dimensional subspaces $V'\leq V$ and $W'\leq W$ together with (finite) bases $\{v_i\}_i$ and $\{w_j\}_j$ that are dual bases to the restrictions of $\{\varphi_i\}_i$ and $\{\psi_j\}_j$ to $V'$ and $W'$, and prove that the kernel element is zero using these bases.
[To add a bit more detail to the last step, suppose that $U$ is a vector space over $k$, and $\alpha_1,\dots,\alpha_d$ a finite linearly independent list of elements of $U^*$. Then the subspace $S=\left\{\left(\alpha_1(u),\dots,\alpha_d(u)\right)\mid u\in U\right\}$ of $k^d$ must be the whole of $k^d$, or else there would be a nonzero linear functional $k^d\to k$ vanishing on $S$, and hence a linear dependence between the $\alpha_i$.
Hence there are elements $u_1,\dots,u_d\in U$ with $\alpha_i(u_j)=\delta_{ij}$ and so $U$ has a finite dimensional subspace $U'=\langle u_1,\dots,u_d\rangle$ with the $u_i$ forming a basis dual to the basis of $(U')^*$ consisting of the restrictions of the $\alpha_i$ to $U'$.]