Let $M,M'$ be homeomorphic smooth, closed, simply connected 4-manifolds. Is it necessarily true that $w_2(TM)=w_2(TM')$ and $p_1(TM)=p_1(TM')$? If so, the comment on this post, shows that $TM$ and $TM'$ are topologically isomorphic as vector bundles.
If the above is false, how does the statement fail, i.e. do we have $w_2(TM)\neq w_2(TM')$, or $p_1(TM)\neq p_1(TM')$, or both?
Yes, at least in the orientable case. First, the Stiefel-Whitney classes only depend on the Wu classes, which are homotopy invariants (they only depend on the cup product + Steenrod operations), and hence so are the Stiefel-Whitney classes, somewhat surprisingly. Second, by the Hirzebruch signature theorem, $p_1$ is determined by the signature (and an orientation), which is determined by oriented homotopy type, and so $p_1$ is a homotopy invariant in this case as well (the dependence on orientation cancels out).
I'm not sure what happens to $p_1$ in the nonorientable case. Maybe we should pass to the orientable double cover.