This line seems especially mistaken: "I think, by definition of a normal subgroup, they are abelian and so this tells us that G is abelian." Certainly normal subgroups need not be abelian: for an example you can take the alternating subgroup of the symmetric group for any $n>5$.
The Sylow theorems tell you that $n_7\in \{1,5,25\}$ and that it is 1 mod 7, and so the only possibility is that it is 1.
The Sylow theorems tell you that $n_5\in \{1,7\}$ and that it is 1 mod 5, and so the only possibility is that it is 1.
Thus for both 5 and 7 you have unique (=normal for Sylow subgroups) subgroups. Let's call them $F$ and $S$ respectively. Clearly $FS$ is a subgroup of $G$ of size 175 by the reasoning you gave. (The reason that $F\cap S$ is trivial is that the intersection is a subgroup of both $F$ and $S$, so it must have order dividing both the order of $F$ and of $S$, but the greatest common divisor is 1.)
$S$ is obviously abelian, as it is cyclic (of prime order!). The question is whether or not a group of size 25 must be abelian. There are a lot of ways to see that, but the one that comes to my mind is to say that it definitely has a nontrivial center. If its center $C$ were of order $5$, then $F/C$ would be cyclic of order 5. However, by a lemma (If $G/Z$ is cyclic for a central subgroup $Z$, then $G$ is abelian) $F$ would have to be abelian.
So $G$ is a product of two abelian subgroups, and so is abelian itself.
And also, your conclusion about the two types of abelian groups of order 175 is correct. Initially you wrote that there were "two isomorphic types," but (I edited that to correct it and ) I hope that was just a slip and that you really did mean "two non-isomorphic types".
To address your second question, all subgroups of order $3$ regardless of what group we are in will be cyclic since all groups of prime order are cyclic.
My hint to you for the first question is the following: Consider the direct product of two groups $G_1 \times G_2$. For some $a \in G_1$ and $b \in G_2$, the order of $(a, b) \in G_1 \times G_2$ will be $lcm(|a|, |b|)$, where $|a|$ denotes the order of $a$ as an element of $G_1$, and likewise for $b$. Of course, you might want to take some time out and prove this fact.
Best Answer
Suppose $G$ is any group with $|G| = 6$. There exist $a\in G$ of order $2$ and $b\in G$ of order $3$ by Cauchy's theorem. If $G$ is abelian, then $ab$ has order $6$, and so $G \cong C_6$. Otherwise, the subgroup $N = \langle b \rangle$ is normal because it has index $2$, and $N$ and $H = \langle a \rangle$ have trivial intersection and satisfy $NH = G$, so $G$ is a semidirect product of $N$ and $H$, which is defined by a nontrivial (since $G$ is not abelian) homomorphism $\phi: H \to \text{Aut}(N) \cong C_2$, of which there is only one. So there are only two groups of order $6$, and only one is abelian, namely $C_6$.
Here's an alternative, more elementary approach. I use Lagrange's theorem, but that's certainly something you'll learn very soon if you haven't already. I'll show there's only one nonabelian group of order $6$, since that seems to be what you really need. (This ends up basically being a proof of the classification, so maybe it's not so great.)
Suppose $G$ is a non-abelian group with $|G| = 6$. By Lagrange's theorem, the order of any element of $G$ is $1$,$2$,$3$, or $6$. If there is an element of order $6$, then $G$ is cyclic and thus abelian, so there is no element of $G$ with order $6$. If every nonidentity element of $G$ has order $2$, then if $a,b\in G$, we have $abab = (ab)^2 = 1$, and thus, multiplying both sides on the right by $ba$, we have $ab = ba$, so $G$ is abelian, and thus there must be some element of $G$ with order $3$. If every nonidentity element has order $3$, then letting $a \in G$ have order $3$, let $b\in G \setminus \{1,a,a^2\}$, and so $b^2 \notin \{1,a,a^2\}$, since otherwise $(b^2)^2 = b$ would be in this set, and it is not. Now let $c \in G \setminus \{1,a,a^2,b,b^2\}$. By similar reasoning, $c^2 \notin \{1,a,a^2,b,b^2\}$, but also $c^2 \ne c$ since $c$ has order $3$. Therefore $G$ contains at least $7$ elements, which is not the case. So there must be some element of $G$ with order $2$.
Let $a\in G$ have order $2$ and $b\in G$ have order $3$ (possible by the above paragraph). You can verify easily that $G = \{1, a, b, b^2, ab, ab^2\}$, as any equality between two of these elements leads quickly to a contradiction (of the facts that $a$ has order $2$ and $b,b^2$ have order $3$), and there are six of them. You can also check that $aba$ is equal to either $b$ or $b^2$ by eliminating the other possibilities in a similar way (equate $aba$ to the other elements in $\{1, a, b, b^2, ab, ab^2\}$ and reach contradictions easily). If $aba = b$, then by multiplying on the right by $a$, we get $ab = ba$, and so $G$ is abelian since $a$ and $b$ generate $G$ and commute. Therefore $aba = b^2$, which implies $ba = ab^2$. You can use this rule to reduce any string of $a$s and $b$s to the form $a^ib^j$ with $0\le i \le 1$ and $0 \le j \le 2$, which is the form in our list $G = \{1, a, b, b^2, ab, ab^2\}$, and so this rule determines the multiplication rule on $G$. Since $a = (12)$ and $b=(123)$ in $S_3$ satisfy these rules, $G$ is isomorphic to $S_3$.