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).
It is not automatic that the order 4 subgroups have only the identity in common - they could (in principle) have an element of order 2 in common.
You have 7 subgroups of order 4 and one of order 7 you know there is at least one element of order 2.
I suggest analysing the orders of the elements you already have, knowing that you have identified everything of order 7 and order 4 (and order 1). How many elements are not in your list? What are the options for the orders of elements which you haven't yet pinned down?
Best Answer
First all elements have order either 1, 2, p or 2p.
If there is an element of order 2p, the group is cyclic, not possible.
If all elements have order 1 or 2, the group is abelian, not possible. Therefore, there is an element of order $p$.
Finally if there is no element of order 2, pick some $x \in G$ and some $y \in G \backslash <x>$. Show that there are $2p-1$ elements in $<x> \cup <y>$.
Pick the remaining element $z$, then $$z^{-1} \in <x> \cup <y> \cup \{ z \}$$
But this is not compatible with all elements having order 1 or p.