Glueing two 2-tangles Given 2-tangles $T_1,T_2\subset B^3$ with their endpoints at some fixed points NW, NE, SW, SE of $\partial B^3$ we can glue them along $\partial B^3$ to obtain a link $L=T_1\cup T_2\subset S^3.$
Q: Does that link $L$ together with $T_1$ determine the homeomorphism class of $(B^3,T_2)$?
Remarks

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*Note that $L$ and $T_1$ do not determine the isotopy class of $T_2$ (within the space of tangles with endpoints at NW, NE, SW, SE, fixed on $\partial B^3$).

*More generally, any 3-manifolds $M_1,M_2$ glued along their boundaries or partial boundaries, $S\hookrightarrow M_1, S\hookrightarrow M_2$, form a 3-manifold $N=M_1\cup_S M_2$.  However $N$ and $M_1$ do not determine the homeomorphism class of $M_2$ in general.

 A: Here's another counterexample.
Let $L$ be the unknot, and $T_1$ the 0-tangle (i.e. the tangle with a diagram consisting of two horizontal strands and no crossings). Then there are lots of possible choice for $T_2$.
For example, $T_2$ can be the doubling $D(J)$ of any knot $J$, where by "doubling" I mean: cut $J$ open to obtain 1-tangle, then take the union of that strand with a parallel copy, to obtain the 2-tangle $D(J)$. This 2-tangle consists of two strands, connecting SW to NW, and SE to NE. Each strand is the knot $J$ cut open.
The $D(J)$ are prime, and $(B^3, D(J))$ is not homeomorphic to $(B^3, D(J'))$ for non-isotopic $J$, $J'$.
A: No, $L$ and $T_1$ together do not determine $T_2$. Take a non-trivial knot $K$ and suppose that $L$ is $K\#K$, and $T_1$ is the trivial tangle. Then $T_2$ can certainly be:

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*a 2-tangle with one boundary-parallel component, and one with a boundary-parallel component connected-summed with $L$;

*a 2-tangle with two boundary-parallel components, each connected-sum with a copy of $K$.

I think you can do similar examples when $L$ and $T_2$ are prime, for instance by cutting a 2-bridge knot in the usual way or in a more "silly" way where all the knotting is in one of the two tangles.
