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**roninpro** To get you started, note that $\displaystyle G$ acts on its Sylow 2-subgroups. If there were 3 of them, then this would induce a homomorphism $\displaystyle \phi:G\to S_3$. On one hand, $\displaystyle \ker \phi$ should not be trivial, because $\displaystyle |S_3|=6$, which is too small. On the other, $\displaystyle \ker \phi$ should not be $\displaystyle G$, because the action is nontrivial. This means that $\displaystyle \ker \phi$ is a proper nontrivial subgroup of $\displaystyle G$. Since the kernel is always normal, this contradicts the simplicity of $\displaystyle G$.

Try using a similar trick for part four.