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$G$ is of order 1225. According to the Sylow theorem, we know that $H$ is the only sylow 5-subgroup of $G$ and $K$ is the only sylow 7-subgroup of $G$. I am trying to prove that they are both normal but looking at the subgroup $gHg^{-1}$ which has the same order of H. What's next shall I look at? And how can we prove that $H$$\cap$$K$ = {e}? Could anyone give me a hint? Thanks!

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For example, $gHg^{-1}$ is also a Sylow $5$-subgroup. There's only one of those. For the second point, the order of $H \cap K$ divides the orders of $H$ and $K$. – Dylan Moreland May 1 '12 at 3:20
A Sylow $p$-subgroup is normal if and only if it is the unique Sylow $p$-subgroup. – Arturo Magidin May 1 '12 at 3:34

Part of the Sylow theorem also says that all p-Sylow subgroups are conjugate. From this you can get that a unique Sylow p-subgroup is normal, as well as characteristic. And, these 3 conditions are equivalent for p-Sylows. For the intersection, as Dylan's comment says, the intersection is a subgroup and must divide the orders of $H$ and $K$.

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