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Let's make an example for Question 1 using classical smoothing theory.

If $K$ is a finite cell complex then for any vector bundle $\xi$ on $K$ you can make a high-dimensional smooth manifold $M$ having a homotopy equivalence $f:M\to K$ such that the tangent bundle is stably $f^\star \xi$. (Embed $K$ in a big $\mathbb R^N$, take a suitable neighborhood, and then take a disk bundle over that.) Furthermore, $M$ is determined by $\xi$ in the sense that if the dimension is big enough and the manifolds are simply connected at $\infty$ then a homotopy equivalence covered by an isomorphism of tangent bundles must be homotopic to a diffeomorphism. (The proof uses the $h$-cobordism theorem.)

Apply this with $K$ being the Moore space $\Sigma^6\mathbb RP^2$. The space of pointed maps $K\to BO$ is the homotopy fiber of the map $\Omega^7BO\to\Omega^7BO$ induced by a degree $2$ map $S^7\to S^7$. Thus, since $\pi_8BO=\mathbb Z$ and $\pi_7BO=0$, there are two vector bundles over $K$ to choose from, stably. This gives, for large $n$, two smooth manifolds of this homotopy type. Their (co)tangent manifolds are diffeomorphic, because they are $2n$-dimensional manifolds of the same homotopy type, both with trivial tangent bundle. (The direct sum of the nontrivial bundle over $K$ with itself is trivial.)

As piecewise linear manifolds $M$ and $M'$ are isomorphic, because the composed maps $K\to BO\to BTop$ are homotopic, because $\pi_8BPL=\mathbb Z$ and $\pi_7BPL=0$, but \pi_7BPL=0$ with the map $\pi_8BO\to \pi_8BPL$ takes taking a generator to an even ($28$) multiple of a generator.

show/hide this revision's text 1

Let's make an example for Question 1 using classical smoothing theory.

If $K$ is a finite cell complex then for any vector bundle $\xi$ on $K$ you can make a high-dimensional smooth manifold $M$ having a homotopy equivalence $f:M\to K$ such that the tangent bundle is stably $f^\star \xi$. (Embed $K$ in a big $\mathbb R^N$, take a suitable neighborhood, and then take a disk bundle over that.) Furthermore, $M$ is determined by $\xi$ in the sense that if the dimension is big enough and the manifolds are simply connected at $\infty$ then a homotopy equivalence covered by an isomorphism of tangent bundles must be homotopic to a diffeomorphism. (The proof uses the $h$-cobordism theorem.)

Apply this with $K$ being the Moore space $\Sigma^6\mathbb RP^2$. The space of pointed maps $K\to BO$ is the homotopy fiber of the map $\Omega^7BO\to\Omega^7BO$ induced by a degree $2$ map $S^7\to S^7$. Thus, since $\pi_8BO=\mathbb Z$ and $\pi_7BO=0$, there are two vector bundles over $K$ to choose from, stably. This gives, for large $n$, two smooth manifolds of this homotopy type. Their (co)tangent manifolds are diffeomorphic, because they are $2n$-dimensional manifolds of the same homotopy type, both with trivial tangent bundle. (The direct sum of the nontrivial bundle over $K$ with itself is trivial.)

As piecewise linear manifolds $M$ and $M'$ are isomorphic, because the composed maps $K\to BO\to BTop$ are homotopic, because $\pi_8BPL=\mathbb Z$ and $\pi_7BPL=0$, but the map $\pi_8BO\to \pi_8BPL$ takes a generator to an even ($28$) multiple of a generator.