$G$ is non abelian simple group of order $<100$ then $G\cong A_5$
By using some more powerful results, it is possible to do this a lot easier.
The two main ingredients for this will be the following:
Burnside's $pq$-Theorem: If only two distinct primes divide the order of $G$, then $G$ is solvable.
Burnside's Transfer Theorem: If $P$ is a $p$-Sylow subgroup of $G$ and $P\leq Z(N_G(P))$ then $G$ has a normal $p$-complement.
So when looking for a non-abelian simple group, the first result immediately tells us that we can assume at least $3$ distinct primes divide the order of $G$.
The way we will use the second result is the following:
Let $P$ be a $p$-Sylow subgroup where $p$ is the smallest prime divisor of $|G|$. We will show that if $|P| = p$ then $P\leq Z(N_G(P))$ (and then the above result says that $G$ is not simple).
To see this, we note that $N_G(P)/C_G(P)$ is isomorphic to a subgroup of $\rm{Aut}(P)$ (this is known as the N/C-Theorem and is a nice exercise), which has order $p-1$, and since the order of $N_G(P)/C_G(P)$ divides $|G|$, this means that $N_G(P) = C_G(P)$ and hence the claim (since we had picked $p$ to be the smallest prime divisor).
The same argument actually shows that if $p$ is the smallest prime divisor of $|G|$, then either $p^3$ divides $|G|$ or $p = 2$ and $12$ divides $|G|$. The reason for this is that if the $p$-Sylow is cyclic of order $p^2$ the precise same argument carries through, since in that case all prime divisors of $|\rm{Aut}(P)|$ are $p$ or smaller.
If $P$ is not cyclic then the order of $\rm{Aut}(P)$ will be $(p^2-1)(p^2 - p) = (p+1)p(p-1)^2$ and the only case where this can have a prime divisor greater than $p$ is when $p = 2$ in which case that prime divisor is $3$, and we get the claim.
In summary we get that the order of $G$ must be divisible by at least $3$ distinct primes, and either the smallest divides the order $3$ times, or the smallest prime divisor is $2$ (actually, by Feit-Thompson, we know this must be the case, but I preferred not to also invoke that), and $3$ must also divide the order.
This immediately gives us $60$ as a lower bound on the order of $G$, and the next possible order would be $2^2\cdot 3\cdot 7 = 84$ and all further orders are greater than $100$. So we are left with ruling out the order $84$ (which you have done), and showing that the only one of order $60$ is $A_5$.
Regarding a group of order $24$.
We know that $n_2$ is either 1 or 3. If $n_2=1$ we are done. If $n_2=3$ then let $X$ be the set of the three $2$-Sylow groups. $G$ acts transitively on $X$ by conjugation. In particular, the action is not trivial. The group action can be thought of as a homomorphism $\phi\colon G\to S_X=S_3$. Since $1<\left|Im(\phi)\right|\le 6$ we have $8\le \left|\ker\phi\right|<24$, so $G$ is not simple.
You can significantly shorten the proof with the following lemma.
If $\left|G\right|=p^km$ where $p$ is a prime, $p\nmid m$ and $p^k\nmid (m-1)!$ then $G$ is not simple.
Proof: Let $P$ be a $p$-Sylow subgroup. Then $G$ acts transitively on the cosets $G/ P$ by left multiplication. Since $\left[G\colon P\right]=m$ this action is a homomorphism $\phi\colon G\to S_m$. If $G$ is simple then $\ker\phi=\{e\}$ so $\phi$ is an injection. That is, $G$ is isomorphic to a subgroup of $S_m$. Hence $p^km=\left|G\right| \mid m!$ so $p^k\mid (m-1)!$ which is a contradiction.
There is a small mistake in your argument, 36 is not a prime squared.