I have two questions that are inspired by a couple of questions here on MO (referenced below), as well as by a conversation with some other grad students at a summer school.

*Caveat*: I'm not a symplectic geometer, nor a differential topologist in the 'classical' sense, so my questions might have a well-known answer, or be open.

It's known that exotic $\mathbb{R}^4$'s have *diffeomorphic* cotangent bundles, as it's known (see here) that the Milnor spheres have *isomorphic* (topological) cotangent bundles.

1. Can thesmoothstructure of cotangent bundles distinguish exotic smooth structures on the base?

Phrased differently, does it exist a pair of smooth manifolds $M,M'$ that are homeomorphic, such that $T^*M$ and $T^*M'$ are isomorphic (see footnote) as vector bundles, but not diffeomorphic as smooth manifolds?

As said before, exotic smooth structures on $\mathbb{R}^4$ are not detected by the cotangent bundle as a smooth manifold. But the cotangent bundle admits a canonical symplectic structure, so...

2. Can thesymplecticstructure of cotangent bundles see the base?

*I.e.* Does it exist a pair $M,M'$ of smooth manifolds that have diffeomorphic tangent bundles, but such that their symplectic cotangent bundles are/are not symplectomorphic? In this phrasing, I'm including also pairs like $\mathbb{R}^3$ and the Whitehead manifold (see this question), but we can impose some further restrictions: what if $M$ and $M'$ are homeomorphic?

**UPDATE**: on this page, Igor Belegradek gives a reference to a paper where it's proved that homeomorphic $n$-spheres have diffeomorphic cotangent bundles. As Andy Putman points out in his answer, Mohammed Abouzaid found examples of spheres with non-symplectomorphic cotangent bundles, so the answer is **YES**: cotangent bundles can (at least sometimes) see the base.

**UPDATE** (16/07/2012): regarding question 2, Tobias Ekholm and Ivan Smith have an interesting preprint: Corollary 1.4 says that the symplectic topology on $T^*(S^1\times S^{8k-1})$ detects the smooth topology of the underlying manifold. Their paper is about double points of Lagrangian immersions (mentioned in Tim Perutz's answer, below).

As Igor Belegradek remarked, the word "isomorphic" might be a bit confusing, in this context. What I mean is that there is a homeomorphism $M\to M'$ that pulls back $T^*M'$ to $T^*M$. This can be strengthened/weakened by asking that this homeomorphism is topologically isotopic to the identity (since $M$ and $M'$ are homeomorphic, this makes sense). For example, I might want $M$ and $M'$ to be both parallelisable.