Galewski-Stern proved
https://mathscinet.ams.org/mathscinet-getitem?mr=420637
" It follows that every topological m-manifold, m≥7 (or m≥6 if ∂M=∅), can be triangulated if and only if there exists a PL homology 3-sphere H3 with Rohlin invariant one such that H3#H3 bounds a PL acyclic 4-manifold."
The Rohlin invariant is a Z/2 valued homomorphsim on the 3-dimensional homology cobordism group, $\Theta_3\to Z/2$, so if it splits there exist non-triangulable manifodls in high dimensions.
I wrote a little expository piece about this and related matters in the Newsletter of the European Mathematical Society:
http://www.ems-ph.org/journals/newsletter/pdf/2010-03-75.pdf
The classical topology of $X:=T^\ast L$ can be taken to include a little more than its diffeomorphism type: there's also an almost complex structure $J$ on $X$, canonical up to homotopy. The pair $(X,J)$ knows the Pontryagin classes of $L$, because $$c_{2k}(TX,J)|_{L}=c_{2k}(TL\otimes\mathbb{C})=(-1)^k p_k(TL),$$
so $(-1)^k p_k(TL)$ pulls back to $c_{2k}(L)$ under projection $X\to L$. However, even with this embellishment, the smooth topology of $X$ doesn't determine $L$.
Faithfulness conjecture: the exact symplectomorphism type of the cotangent bundle $(X,\omega=d\lambda_L)$ of a compact manifold $L$ determines $L$.
An exact symplectomorphism $T^\ast L \to T^\ast L'$ is a diffeomorphism $f$ such that $f^*\lambda_{L'}-\lambda_L= dh$ for $h$ a compactly supported function. The conjecture (but not the name) is standard.
Attempts to use symplectic invariants of $X$ to distinguish smooth structures on $L$ have so far been a complete failure. For example, the symplectic cohomology ring $SH^*(X)$ is isomorphic to loopspace homology $H_{-*}(\mathcal{L}L; w)$ (the coefficients are the local system $w$ of $\mathbb{Z}$-modules determined by $w_2$), with the string product. This invariant is determined by the homotopy type of $L$.
"Arnol'd's conjecture" (scare quotes because Arnol'd really made a much more circumspect conjecture). Any exact Lagrangian embedding $\Lambda \to X$ (with $\Lambda$ compact) is exact-isotopic to the embedding of the zero-section.
This would immediately imply the faithfulness conjecture.
There has been progress towards Arnol'd's conjecture of three kinds:
(1) It's true for $L=S^2$ (Hind, http://arxiv.org/abs/math/0311092).
(2) The work of several authors (Fukaya-Seidel-Smith http://arxiv.org/abs/0705.3450, Nadler http://arxiv.org/abs/math/0612399, Abouzaid http://arxiv.org/abs/1005.0358, and Kragh's work in progress) cumulatively shows that the projection from an exact Lagrangian to the zero-section is a homotopy equivalence. This is good evidence for the truth of the conjecture, but for the application to faithfulness one might as well make homotopy-equivalence an assumption.
(3) As Andy mentioned, Abouzaid http://arxiv.org/abs/0812.4781 has shown that a homotopy $(4n+1)$-sphere $S$, such that $T^*S$ contains an exact embedded Lagrangian $S^{4n+1}$, bounds a parallelizable manifold. This is proved by a stunning analysis of the geometry of a space of pseudo-holomorphic discs.
The existence of exact Lagrangian immersions is governed by homotopy theory (there is an h-principle which finds such an immersion given suitable homotopical data). Just as the subtleties of 4-manifold topology can be located at the impossibility of removing double points of immersed surfaces (the failure of the Whitney trick), so the subtlety of the symplectic structure of cotangent bundles comes down to the question of removability of double points of Lagrangian immersions.
Best Answer
For a pair of smooth, simply connected, compact, oriented 4-manifolds $X$ and $Y$,
Any isomorphism of the intersection lattices $H^2(X)\to H^2(Y)$ comes from an oriented homotopy equivalence $Y\to X$ (Milnor, 1958).
Any oriented homotopy equivalence is a tangential homotopy equivalence (Milnor, Hirzebruch-Hopf 1958).
Any oriented homotopy equivalence comes from an h-cobordism (Wall 1964).
Any oriented homotopy equivalence comes from a homeomorphism (Freedman).
It need not be the case that $X$ and $Y$ are diffeomorphic (Donaldson). Many examples are now known: e.g., Fintushel-Stern knot surgery on a K3 surface gives a family of exotic K3's parametrized by the Alexander polynomials of knots.
Here's a sketch of why homotopy equivalences preserve tangent bundles: $X$ and $Y$ have three characteristic classes: $w_2$, $p_1$ and $e$. However, $e[X]$ is the Euler characteristic, and $p_1[X]$ three times the signature. By the Wu formula, $w_2$ is the mod 2 reduction of the coset of $2H^2(X)$ in $H^2(X)$ given by the characteristic vectors, hence is determined by the lattice. In trying to construct an isomorphism of tangent bundles over a given homotopy equivalence, the obstructions one encounters are in $H^2(X;\pi_1 SO(4))=H^2(X;Z/2)$ and in $H^4(X;\pi_3 SO(4))=Z\oplus Z$, and these can be matched up with the three characteristic classes.