This is to shed some light on Part 3 which asks for a classification of normal vector bundles of a smooth $n$-dimensional submanifold $X$ of $\mathbb R^{2n}$.
The normal bundle $\nu$ to $X$ is stably isomorphic to the negative of $TX$, and in particular the Pontryagin and Stiefel-Whitney classes of $\nu$ are completely determined by those of $X$. Also if $X$ is orientable, then the Euler class of $\nu$ vanishes (contactibility of $\mathbb R^{2n}$ allows to push the embedding off itself). On the other hand, if $X$ is non-orientable, the twisted Euler class (i.e., the first obstruction to the existence of a nowhere zero section) of $\nu$ can be nonzero and is an important invariant (see below).
For simplicity let us assume that $X$ is closed. Let us also ignore the cases $n\le 3$ where if $X$ is orientable one can use the classification of oriented bundles Dold-Whitney in [Classification of Oriented Sphere Bundles Over A 4-Complex], and presumably with some work one deal with the nonorientable case.
Instead of classifying normal bundles to $X$ let us describe isotopy classes of embeddings of $X$ into $\mathbb R^{2n}$. Of course, isotopic embeddings have isomorphic normal bundles. One quick statement is a theorem of Haefliger-Hirsch [On the existence and classification of differentiable embeddings. Topology 2 1963 129–135] that
If $X$ is simply-connected and $n\ge 4$, then there is only one isotopy class of embeddings from $X$ into $\mathbb R^{2n}$, and hence, only one normal bundle.
More generally, if $X$ is orientable and $n\ge 4$, then the set of isotopy classes of smooth embeddings of $X$ into $\mathbb R^{2n}$ is bijective to $H_1(X)$ if $n$ is odd, and to $H_1(X;\mathbb Z_2)$ if $n$ is even.
If $X$ is non-orientable and $n$ is even and $\ge 4$, there is a more complicated classification by Kitada in [Classification of embeddings of a non-orientable manifold. J. Fac. Sci. Univ. Tokyo Sect. IA Math. 18 (1971/72), 435–442]. The conclusion is that the set of isotopy classes is bijective to $\mathbb Z\oplus H^{n-1}(X)/K$ where $H$ is a certain subgroup of $H^{n-1}(X)$. The $\mathbb Z$-factor corresponds to the twisted Euler class of $\nu$, and in particular, if $H^{n-1}(X)=0$, then $\nu$ is completely determined by the twisted Euler class.
I don't know what happens if $X$ is non-orientable and $n$ is odd.