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. 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.