If you are assuming that $G$ is an abelian group of order $100$, then you don't need Sylow's Theorems, you just need Cauchy's Theorem: since $2$ and $5$ divide $|G|$, $G$ has an element $a$ of order $2$, and an element $b$ of order $5$. Then $\langle a\rangle\cap\langle b\rangle = \{0\}$ (writing the groups additively), so $a+b$ has order $10$, as is easily verified:
$$\begin{align*}
k(a+b) = 0 &\iff ka+kb = 0\\
&\iff ka=-kb\\
&\iff ka,kb\in\{0\}\\
&\iff 2|k\text{ and }5|k\\
&\iff 10|k.
\end{align*}$$
The second part asks you to whether there are no elements of order greater than 100 in an abelian group of order 100? The cyclic group of order 100 shows that this need not be the case. In fact, the only abelian group of order $100$ in which there are no elements of order greater than $10$ is the group $\mathbf{Z}_2\oplus\mathbf{Z}_2\oplus\mathbf{Z}_{5}\oplus\mathbf{Z}_{5}\cong \mathbf{Z}_{10}\oplus\mathbf{Z}_{10}$.
Or I may be misunderstanding the seecond part, and instead you are told that $G$ has no elements of order greater than $10$... the only possible orders, by Lagrange's Theorem, are $1$, $2$, $4$, $5$, $10$, $20$, $25$, $50$, and $100$. Since you are told there are no elements of order greater than $10$, then the orders must be $1$, $2$, $4$, $5$, or $10$. But if you have an element $x$ of order $4$, then $x+b$ is of order $20$ (same argument as above), a contradiction. So every element is of order $1$, $2$, $5$, or $10$. And there are certainly elements of order $1$ (namely, $0$), order $2$ and $5$ (Cauchy's Theorem), and order $10$ (first part of the problem).
Note that in fact any abelian subgroup of index $p$ must contain the center: for if $H$ is maximal and abelian, and $Z(G)$ is not contained in $H$, then $G=HZ(G)$, which would make $G$ abelian. Therefore, if $H$ and $K$ are both abelian of index $p$, then $Z(G)\subseteq H\cap K$. If you have proven that $H\cap K\subseteq Z(G)$, then in fact we have that $H\cap K=Z(G)$ for any two distinct abelian subgroups of index $p$.
(And indeed: if $H\neq K$ are both abelian of index $p$, then for every $g\in G$ there exists $h\in H$, $k\in K$ with $g=hk$; given $x\in H\cap K$ we have $gx = (hk)x = x(hk) = xg$, since $x$ commutes with everything in $K$ and with everything in $H$, so $H\cap K\subseteq Z(G)$).
Conversely, if $M$ is a subgroup of $G$ that contains $Z(G)$ and is of index $p$, then it must be abelian (since $M/Z(G)$ is of order $p$). Thus, every abelian subgroup of $G$ of index $p$ corresponds to a subgroup of $G/Z(G)$, and you are done.
For part (b), show that (a) implies that if $G$ has more than one abelian subgroup of index $p$, then $G$ is of class exactly $2$: by the argument in (a), you know that $Z(G)$ has index $p^2$ in $G$, but that means that $G/Z(G)$ is abelian, which means that $[G,G]\subseteq Z(G)$.
So if $G$ is of class $3$ and order $16$, then it has at most one abelian subgroup of order $8$. The center must be of order $2$: if the center were of order $4$, then $G/Z(G)$ would be of order $4$, hence abelian, so again we have that $G$ would be of class $2$. Thus, $G/Z(G)$ is of order $8$, and cannot be abelian. Therefore, $G/Z(G)$ is one of the two nonabelian groups of order $8$. If $G/Z(G)$ were quaternion, then it would have four cyclic subgroups of order $4$: each of them pull back to a subgroup $H$ of $G$ of order $8$, containing the center; and it is not hard to show that they are all abelian, contradicting the fact that $G$ has at most one abelian subgroup of order $8$. So $G/Z(G)$ must be dihedral. Verify that you get a cyclic subgroup of order $8$ in $G$ in this case.
Added. So, let us suppose that $G$ is of class $3$ and order $16$, and has a unique abelian subgroup of index $2$. We know $Z(G)$ is of order $2$, and that $G/Z(G)$ is dihedral of order $8$. Let $z\in G$ generate $Z(G)$, and let $x\in G$ map to a generator of the cyclic group of order $4$ in $G/Z(G)$. We know that $x$ has order either $4$ or $8$ in $G$. We aim to show that it has order $8$.
If $x$ has order $8$, we are done. Otherwise, $|x|=4$. Let $y\in G$ be such that $y$ maps to the generators of $G/Z(G)$ of order $2$, so that $yx = x^3y$ or $yx=x^3yz$. In the former case, we have $yx^2 = x^6y = x^2y$, so $x^2$ commutes with $x$, $y$, and $z$; in the latter case we would have $yx^2 = x^3yxz = x^6yz^2 = x^2y$, so again $x^2$ commutes with $x$, $y$, and $z$. Since $x$, $y$, and $z$ generate $G$, it follows that $x^2\in Z(G)$. But then the image of $x$ in $G/Z(G)$ would have order $2$, which is a contradiction. Therefore, $x$ cannot have order $4$, and so must have order $8$, as desired.
Best Answer
The Heisenberg group modulo $5$ is a non-abelian group of order $125$.
It is the set of all $3 \times 3$ matrices with entries in $\mathbb F_5$ of the form $$ \begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix} $$