$\require{AMScd}\newcommand{\id}{\operatorname{id}}$I use a common characterisation of the approximation property as found in e.g. Ryan's book Zbl 1090.46001.

A Banach space $X$ has the approximation property if and only if for each Banach space $Y$ (it is enough to take $Y=X^*$) the natural map
$$ X \widehat\otimes Y \rightarrow X \check\otimes Y $$
is injective.

Here I write $\widehat\otimes$ and $\check\otimes$ for the completed projective, respectively, injective tensor products.

We can now answer (2) in the negative. Let $X$ have the approximation property, and let $S:W\rightarrow V$ be injective. Consider the commutative diagram
$$ \begin{CD}
X\widehat\otimes W @>>> X \check\otimes W \\
@V{\id\otimes S}VV @VV{\id\otimes S}V \\
X\widehat\otimes V @>>> X \check\otimes V
\end{CD} $$
The map $\id\otimes S: X \check\otimes W \rightarrow X \check\otimes V$ is injective, and the horizontal arrows are injective as $X$ has AP, so $\id\otimes S: X \widehat\otimes W \rightarrow X \widehat\otimes V$ is injective. In particular $X=L_1$ has the AP, showing the negation of (2).

As Jochen Wengenroth noted, Q1 can be reduced to the $T\otimes S$ case which the OP stated has a positive answer. However, here is a concrete example, following Chapter 5, Corollary 4 of Defant and Floret Zbl 0774.46018. Let $X$ be any Banach space, and let $B_{X^*}$ be the unit ball of the dual space $X^*$, consider $\ell_\infty(B_{X^*})$ and define $j:X\rightarrow \ell_\infty(B_{X^*})$ by evaluation: $j(x) = ( \phi(x) )_{\phi\in B_{X^*}}$. Then $j$ is an isometry onto its range. We know that $\ell_\infty(B_{X^*})$ has AP so
$$ X^* \widehat\otimes \ell_\infty(B_{X^*}) \rightarrow X^* \check\otimes \ell_\infty(B_{X^*}) $$
is injective. Consider now the commutative diagram
$$ \begin{CD}
X^* \widehat\otimes X @>>> X^* \check\otimes X \\
@V{\id\otimes j}VV @VV{\id\otimes j}V \\
X^* \widehat\otimes \ell_\infty(B_{X^*}) @>>> X^* \check\otimes \ell_\infty(B_{X^*}) \\
\end{CD} $$
The bottom arrow is injective, and the right-hand down arrow is. If $X$ does not have AP then the top arrow is not injective, and so the left-hand
down arrow must fail to be injective, which gives an example of (1). (There is nothing special about $\ell_\infty$ here: any Banach space $F$ with the AP and any injection $j:X\rightarrow F$ would work.)