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).
You already did 70% of the work:
Let $G$ be of order $45 = 5\cdot 3^2$. By Sylow's second theorem, the number of 5-sylow-groups $n_5$ must be either 1, 3, or 9. Out of those choices, only $n_5=1\equiv 1$ modulo 5. Therefore, only one subgroup of order 5 exists (normality isn't even important here).
Now note that every possible element of order five generates a subgroup of order five (in which it must be contained). I'm confident you can take it from here. :-)
Best Answer
Here is a sketch proof. Let $G$ be a group of order $21$. Then by Cauchy $G$ has an element $x$ of order $7$. Let $P = \langle x \rangle$. Then $P$ is the unique subgroup of $G$ of order $7$, becuase if there was another one $Q$ then we would have $|PQ|=49>|G|$. So $P \lhd G$.
By Cauchy again, $G$ has an element $y$ of order $3$. Then the elements of $S = \{ x^iy^j : 0 \le i \le 6, 0 \le j \le 2 \}$ are all distinct, so $|S|=21$ and hence $S=G$.
Since $P \lhd G$, we have $yxy^{-1} = x^i$ for some $i$ with $1 \le i \le 6$. So $yx = x^iy$, and this relation together with $x^7=y^3=1$ enable us to multiply any two elements of $S$ together. So the isomorphism type of $G$ is determined by $i$.
These three relations also imply that $i=1,2$ or $4$ (I'll leave that as an exercise). Note that $yxy^{-1} = x^4 \Rightarrow y^{-1}xy = x^2$, so by interchanging $y$ and $y^{-1}$, we see that the groups defined by $i=2$ or $4$ (assuming that they exist at all) are isomorphic.
So there are at most two isomorphism types of groups of order $21$, which was what you had to prove. In fact, when $i=1$, we get the abelian group $C_7 \times C_3$, and when $i=2$ or $4$, we get the nonabelian group.