I am somewhat a beginner in the field of operator algebras and was wondering about the following:

Let $T$ be a linear map between the space of bounded operators $B(H)$ on some Hilbert space and $S$ a map between the space of trace-class operators that we denote by $N(H)$ in the sequel.

Then, one defines maps $T_n: M_n(B(H)) \rightarrow M_n(B(H))$ by $$T_n((a_{ij})):=(T(a_{ij}))$$

Then, $T$ is completely bounded if and only if $\sup_n \left\lVert T_n \right\rVert< \infty.$

And analogously for $S.$

I would like to know whether a map $T$ is completely bounded if and only if $$\left\lVert T \otimes \operatorname{id}_{X} \right\rVert< \infty$$ or in case of $S$ whether complete boundedness is equivalent to $$\left\lVert S \otimes \operatorname{id}_{X} \right\rVert< \infty$$

for some $X$?

For $T$ I think I found a reference on math.stackexchange saying that $X=B(K)$ works for any infinite-dimensional Hilbert space $K$ (nothing on whether $K$ needs to be separable discussed in this thread.) This should imply that $X=N(H)$ works for $S$ as well.

Assuming this to be true, I am wondering about two things now:

1.)Assume I can take $X=B(H)$ for $T$ and $X=N(H)$ for $S$, then $T \otimes \operatorname{id}_{B(H)}$ is a map from $B(H) \otimes B(H)$ into itself. Can the space $B(H) \otimes B(H)$ be identified with $B(H \otimes H)$?- I assume that by a duality argument this would be equivalent to asking whether $N(H) \otimes N(H)$ is isomorphic to $N(H\otimes H)$.

2.) Are there any other choices of $X$ allowed in the above two examples?