Here is my comment with some added detail:
Let $G$ be a group of order $182$. You already concluded that you have a normal $7$-Sylow subgroup, so let us call this subgroup $P$ and look at $G/P$.
$G/P$ has order $26$ so the $13$-Sylow subgroup has index $2$. This gives us a subgroup of $G$ of index $2$, which is normal due to the index. We will call this subgroup $N$. So now we have that $G$ is a semidirect product of $N$ and a $2$-Sylow subgroup which we will call $Q$.
Let us consider $N$ a bit more carefully. It is a group of order $91 = 7\cdot 13$ and we easily see from the Sylow theorems that both its Sylow subgroups are normal, and hence $N$ is cyclic.
So we now know that our group is a semidirect product of a cyclic group of order $91$ with one of order $2$, so we need to look for subgroups of $\rm{Aut}(C_{91})$ of order $1$ or $2$ (the one of order $1$ just gives us a direct product, which will give the cyclic group of order $182$ which is the only abelian possibility).
Now, $\rm{Aut}(C_{91})$ is isomorphic to $C_6\times C_{12}$ so it has $3$ distinct subgroups of order $2$ and we need to determine which of these yield distinct semidirect products.
To do this, it is probably a good idea to see precisely how these automorphisms of order $2$ act on $C_{91} \simeq C_7\times C_{13}$. The three possible automorphisms are then $(g,h)\mapsto (g^{-1},h)$, $(g,h)\mapsto (g,h^{-1})$ and $(g,h)\mapsto (g^{-1},h^{-1})$.
To see that these give three distinct semidirect products, we can count the number of elements of order $2$ in each.
An arbitrary element can be written as $(g,h)\varphi$ where $\varphi$ is either the identity or the given automorphism of order $2$. Clearly if $\varphi$ is the identity, the element cannot have order $2$, so let us now compute the square of such an element in each of the three cases (remember that $\varphi$ is its own inverse and that conjugation by $\varphi$ corresponds to applying $\varphi$).
In the first case, we get $(g,h)\varphi(g,h)\varphi = (g,h)(g^{-1},h) = (1,h^2)$ which is the identity if and only if $h^2 = 1$ which means if and only if $h = 1$, so we get precisely $7$ elements of order $2$ in this group.
Similarly, in the second case, we get that the element has order $2$ if and only if $g = 1$ so we get $13$ element of order $2$ in this case.
In the final case, we see that all elements of the given form have order $2$, so we have $91$ such elements (this case gives us the dihedral group).
Best Answer
Hints:
1) Prove there are exactly one Sylow 5-subgroup and one Sylow 7-subgroup
2) Prove that every group of order the square of a prime number is abelian
3) Show thus that any group of order $\,1225\,$ is abelian and thus
4) There are only $\,4\,$ groups of order $\,1225\,$ (Using the Fundamental Theorem of Finitely Generated Abelian Groups)