As YCor notes in comments, (2) is a special case of (1), so I'll only address (1).

Suppose $\Lambda^k\varphi:\Lambda^kV\to\Lambda^kW$ is not injective, and let $x\neq0$ be in the kernel.

Then $x$ can be written in the form
$$x=\sum_{i=1}^mv_{i1}\wedge\dots\wedge v_{ik}.$$
Let $V'$ be the finite dimensional subspace of $V$ spanned by $\{v_{ij}\mid1\leq i\leq m,1\leq j\leq k\}$, and $\varphi':V'\to W$ the restriction of $\varphi$ to $V'$.

Let
$$x'=\sum_{i=1}^mv_{i1}\wedge\dots\wedge v_{ik},$$
considered as an element of $\Lambda^kV'$. Then $x'\neq0$, since it is sent to $x$ by the map induced by the inclusion of $V’$ into $V$. Also
$$\Lambda^k\varphi'(x')=\Lambda^k\varphi(x)=0,$$
so $x'$ is a nonzero element of the kernel of $\Lambda^k\varphi'$.

Hence we may as well assume that $V$ is finite dimensional.

The fact that $\Lambda^k\varphi(x)=0$ follows from a finite number of the relations defining the exterior power $\Lambda^kW$, involving only finitely many elements of $W$. If we replace $W$ by the finite dimensional subspace $W''$ spanned by the image of $\varphi$ and these finitely many elements, then $\varphi$ induces a map $\varphi'':V\to W''$, and $\Lambda^k\varphi''(x)=0$, since the same relations that implied $\Lambda^k\varphi(x)=0$ in $\Lambda^kW$ also imply that $\Lambda^k\varphi''(x)=0$ in $\Lambda^kW''$.

Hence we can also assume that $W$ is finite dimensional, and what remains is a problem about finite dimensional vector spaces that can easily be answered without choice by choosing bases.

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