It is clear that a finte group of order $105$ is not simple since it contains a normal Sylow $7-$subgroup or a normal Sylow $5-$subgroup. I wonder if it is possible that there exists a Sylow $7-$subgroup and a Sylow $5-$subgroup ?
Of course there exist Sylow $7$-subgroups and Sylow $5$-subgroups: Sylow's Theorems tell you that the group has Sylow $p$-subgroups for every $p$ (with the Sylow $p$-subgroup being trivial if $p$ does not divide $105$).
If you meant to ask "Can the group have normal Sylow $5$- and Sylow $7$ subgroup?" then the answer is "yes": the (unique) abelian group of order $105$ will do (since $105=3\times 5\times 7$ is square free, the only abelian group of order $105$ is the cyclic one).
In general, you can always find a finite group of order $n$ in which all Sylow subgroups are normal: any abelian group of order $n$, and there is always at least one, will do. If any of the primes in the factorization of $n$ is raised to a degree greater than $2$, then you can also find a nonabelian group in which all the Sylow subgroups are normal, just by taking the direct product of $p$-groups of the correct size, and picking a nonabelian one when the prime is raised to a sufficiently large degree.
And in this case, you can get a nonabelian group in which the Sylow $7$- and the Sylow $5$-subgroups are normal by first constructing a nonabelian group of order $21$ (there are no nonabelian groups of orders $15$ or $35$) which has a normal subgroup of order $7$, and then taking the direct product with with $C_5$.
For any group of order 105, there always exists a normal sylow 7-subgroup and a normal sylow 5-group.