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 $U'$.]

In fact, answering a question asked in comments, there is a common generalization of (1) and (2). For any vector spaces $V$, $V'$, $W$, $W'$, the fact that the natural map
$$\text{Hom}(V,V')\otimes\text{Hom}(W,W')\to\text{Hom}(V\otimes W,V'\otimes W')$$
is injective can be proved without the axiom of choice.

To see this, note that an element of the kernel can be written in terms of $\alpha_i\otimes\beta_j$, where $\alpha_1,\dots,\alpha_s$ are finitely many linearly independent elements of $\text{Hom}(V,V')$ and $\beta_1,\dots,\beta_t$ finitely many linearly independent elements of $\text{Hom}(W,W')$.

I claim that there is a finite dimensional subspace $V''\leq V$ such that the restrictions of $\alpha_1,\dots,\alpha_s$ to $V''$ are still linearly independent (and similarly for $W$ and the $\beta_j$). This follows by induction on $s$. Suppose there is a finitely generated subspace $U$ so that the restrictions of $\alpha_1,\dots,\alpha_k$ to $U$ are linearly independent. Then either the restrictions of $\alpha_1,\dots,\alpha_{k+1}$ to $U$ are linearly independent, or there is a linear dependence $\sum_{i=1}^{k+1}\lambda_i(\alpha_i|_U)=0$, which is unique up to a scalar multiple. But since $\alpha_1,\dots,\alpha_{k+1}$ are linearly independent, there is some $v\in V$ such that $\sum_{i=1}^{k+1}\lambda_i\alpha_i(v)\neq0$, and then the restrictions of $\alpha_1,\dots,\alpha_{k+1}$ to $U+\langle v\rangle$ are linearly independent.

Replacing $V$ and $W$ by $V''$ and $W''$, and $V'$ and $W'$ by $\sum_i\alpha_i(V'')$ and $\sum_j\beta_j(W'')$ reduces the original question to one about finite dimensional vector spaces, which can easily be solved without invoking choice.