= Solution
For order $48=2^4\cdot3$, the <Sylow theorems> give $n_2\mid3$ and $n_2\equiv1\pmod2$, hence $n_2=1$ or $3$. If $n_2=1$, the <Sylow subgroup> of order $16$ is a nontrivial proper <normal subgroup>. If $G$ were simple in the other case, it would be nonabelian (a finite abelian <simple group> has prime order), and the preceding result would force $48\mid3!/2=3$, impossible. Thus \b[every group of order 48 is not simple].
An example with no normal <Sylow subgroup> for $p=2$ is
$$
\boxed{G=S_4\times C_2}.
$$
In $S_4$, $P=\langle(1234),(13)\rangle$ is a <dihedral group> of order $8$: conjugation by $(13)$ inverts $(1234)$. Thus $P\times C_2$ has order $16$ and is a <Sylow subgroup> of $G$. The only transpositions in $P$ are $(13)$ and $(24)$; conjugating $(13)$ by $(23)$ gives $(12)\notin P$. Therefore $P$ and $P\times C_2$ are not normal. By conjugacy, no other <Sylow subgroup> for $p=2$ is normal either.
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