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?
As was remarked before, you do not have to assume that $G \cong A_5$.
(1) You did that one correctly!
(2) If $|Syl_3(G)|=4$, and $S \in Syl_3(G)$, then $|G:N_G(S)|=4$. Now let $G$ act on the left cosets of $N_G(S)$ by left multiplication, then the kernel of this action is $core_G(N_G(S))=\bigcap_{g \in G}N_G(S)^g$, which is a normal subgroup. Hence, by the simplicity of $G$, it must be trivial and $G$ can embedded in $S_4$, a contradiction, since $60 \nmid 24$.
(3) We prove that if a non-abelian simple $G$, with $|G|=60$, has an abelian subgroup of order $6$, then $G \cong A_5$. This gives a contradiction, since it is easily seen that $A_5$ does not contain any elements of order $6$ (note that an abelian group of order $6$ must be cyclic).
So assume $H \lt G$ is abelian and $|H|=6$. $H$ is not normal so, $N_G(H)$ is a proper subgroup (if not then $H$ would be normal) and since $|G:N_G(H)| \mid 10$, we must have $|G:N_G(H)|=5$ ($=2$ is not possible since subgroups of index $2$ are normal). Similarly as in (2), $G/core_G(N_G(H))$ embeds homomorphically in $S_5$ this time. Of course $core_G(N_G(H))=1$, so $G$ is isomorphic to a subgroup of $S_5$ and since it is simple it must be isomorphic to $A_5$ (if we write also $G$ for the image in $S_5$, consider $G \cap A_5 \lhd G$ and use $|S_5:A_5|=2$).
(4) In general: if $G$ is a group with a unique element $x$ of order $2$, then $x \in Z(G)$. Why? Because for every $g \in G$, $g^{-1}xg$ has also order $2$ and must be equal to $x$. In your case $G$ is non-abelian simple, so $Z(G)=1$.
So only (1) is the true statement.
Edit For case (3) I forgot the case where $|G:N_G(H)|=10$. I have a proof that is quite sophisticated and maybe there is an easier way.
Anyway, in this case $H=N_G(H)$. Consider the subgroup $P$ of order $3$ of $H$. This must be a Sylow $3$-subgroup of $G$, since $3$ is the higest power of $3$ dividing $|G|=60$. Observe that in fact $N_G(P)=H$. This follows from what we showed in (2): $|Syl_3(G)|=|G:N_G(P)|=10$ and of course $H \subseteq N_G(P)$.
Trivially, $P \subset Z(N_G(P))$. Now $P$ satifies the criterion of Burnside's Normal $p$-Complement Theorem, see for example Theorem (5.13) here. But then $P$ has a normal complement $N$, such that $G=PN$ and $P \cap N=1$. Now $G$ is non-abelian simple, so $N=1$ or $N=G$, which both lead to a contradiction.
Best Answer
It is a much more general phenomena:
Claim: If $|H|=n$ and $|K|=m$ and $\gcd(m,n)=1$ then $H\cap K=\{e\}$.
Proof: Let $g\in H\cap K$. Then the order of $g$, $|g|$ divides both $n$ and $m$ (why?) and hence $|g|$ divides $1$ - so $g=1$.