To answer your question 2, there are very few pairs of finite simple groups with the same order. There's $PSL_3(4)$ and $PSL(4,2)$, and there's $P\Omega_{2n+1}(q)$ and $PSp_{2n}(q)$ for $q$ odd. None of these are Sylow isomorphic in your sense.

As a contribution to your question 1, finite groups can be not only Sylow isomorphic but they can even be indistinguishable from the point of view of fusion at any prime. This is equivalent to the plus construction on the classifying spaces being homotopy equivalent. Probably the smallest example is $(\mathbb{Z}/3\rtimes\mathbb{Z}/4) \times (\mathbb{Z}/5\rtimes\mathbb{Z}/2)$ and $(\mathbb{Z}/3\rtimes\mathbb{Z}/2) \times (\mathbb{Z}/5\rtimes\mathbb{Z}/4)$, with the action of $\mathbb{Z}/4$ having the subgroup of index two in its kernel in both cases.

To answer your question 2, there are very few pairs of finite simple groups with the same order. There's $PSL_3(4)$ and $PSL(4,2)$, and there's $P\Omega_{2n+1}(q)$ and $PSp_{2n}(q)$ for $q$ odd. None of these are Sylow isomorphic in your sense.

As a contribution to your question 1, finite groups can be not only Sylow isomorphic but they can even be indistinguishable from the point of view of fusion at any prime. This is equivalent to the $\mathbb{Z}$-completion of the classifying spaces being homotopy equivalent. Probably the smallest example is $(\mathbb{Z}/3\rtimes\mathbb{Z}/4) \times (\mathbb{Z}/5\rtimes\mathbb{Z}/2)$ and $(\mathbb{Z}/3\rtimes\mathbb{Z}/2) \times (\mathbb{Z}/5\rtimes\mathbb{Z}/4)$, with the action of $\mathbb{Z}/4$ having the subgroup of index two in its kernel in both cases.