I want a reference of the following statement which I think is true. Let $X$ and $Y$ be Banach spaces with $X$ finite dimensional. Then $(X\otimes_\epsilon Y)^*$ is isometrically isomorphic to $(X^*\otimes_\pi Y^*)$ and $(X\otimes_\pi Y)^*$ is isometrically isomorphic to $(X\otimes_\epsilon Y)^*$ where $\otimes_\epsilon$ and $\otimes_\pi $ denote the injective and projective tensor product of Banach spaces respectively. For tensor product of Banach spaces see the book https://www.amazon.in/Raymond-A-Ryan/e/B001K88BH0/ref=dp_byline_cont_pop_ebooks_1.
Functional Analysis – Duality of Projective and Injective Tensor Product
banach-spacesfa.functional-analysisoperator-theory
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$\require{AMScd}\newcommand{\id}{\operatorname{id}}$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.)
Best Answer
Since Ryan's book is mentioned in the original question, here's some pointers for an answer based on this source:
For the dual of $X\otimes_\pi Y$ see Section 2.2 where it's shown that $(X\otimes_\pi Y)^* = B(X,Y^*)$ the space of all bounded linear maps. We always have that $X^* \hat\otimes_\epsilon Y^*$ is a subspace of $B(X,Y^*)$, see Section 3.1, and it's easy to see that when $X$ is finite-dimensional, then we have equality here.