= Solution
Work in the finite-group setting. If $M$ is maximal in a nonabelian simple $G$, then $M\ne1$: otherwise every nonidentity element would generate $G$, forcing a cyclic group of prime order. Choose a <minimal normal subgroup> $N\ne1$ of $M$. Every <characteristic subgroup> of $N$ is normal in $M$, so minimality makes $N$ a <characteristically simple group>. Its <normalizer> contains $M$. By maximality it is $M$ or $G$, and the latter would make $N$ a nontrivial <normal subgroup> of the <simple group> $G$, impossible because $N\le M<G$. This proves the <maximal subgroup normalizer criterion>:
$$
\boxed{M=N_G(N)\text{ for some characteristically simple }N.}
$$
For $A_5$, <stabilizer subgroups> have order 12. <Stabilizer subgroups> of a two-element subset have order $|(S_2\times S_3)\cap A_5|=6$, and are isomorphic to $S_3$; for example $\langle(1\,2\,3),(1\,2)(4\,5)\rangle$. The <normalizer> of a Sylow 5-subgroup has order 10: its <normalizer> in $S_5$ has order $5\cdot4$, and exactly half its <permutations> are even, since a multiplier of order four on $\mathbb F_5$ is an odd four-cycle. This gives a <dihedral group> of order 10.
To prove completeness, an intransitive <subgroup> either fixes a point, or its orbits on five letters have sizes two and three. It therefore lies in one of the first two types. A transitive proper <subgroup> $H$ has order divisible by 5. Its number of Sylow 5-subgroups is one or six, since $A_5$ has six and the count is $1$ modulo 5. A count of six would make $30\mid |H|$, forcing an index-two <subgroup> and contradicting <Simplicity of the alternating group A5>. Therefore $H$ normalizes its unique Sylow 5-subgroup and lies in the order-ten <normalizer>.
<Stabilizer subgroups> are maximal because their index is prime. The order-ten group is transitive and cannot have a larger proper transitive overgroup by the preceding argument. The order-six group has no fixed point, cannot lie in a <stabilizer subgroup>, and cannot lie in the order-ten type because 3 does not divide 10. Thus all three types are maximal. $A_5$ is transitive on points and on unordered pairs, while Sylow conjugacy handles the <normalizers>. Consequently
$$
\boxed{A_5:\text{ three maximal-subgroup classes, of orders }6,\ 10,\ 12.}
$$
Now let $G=GL_3(2)$. Choosing independent columns gives $|G|=7\cdot6\cdot4=168$. Its simplicity follows from the elementary projective-transvection proof in part 4(i), which does not use this <subgroup> classification.
A <stabilizer subgroup> and a plane <stabilizer subgroup> each have index seven and order 24. To construct the remaining type, identify $\mathbb F_2^3$ with $\mathbb F_8$ and take the <Singer cycle> $P$ of multiplications by nonzero field elements, of order seven. Any <linear map> commuting with a primitive multiplication commutes with all of $\mathbb F_8$, so its <centralizer> is precisely $P$. A <normalizer> sends a primitive element $a$ to another root of its minimum polynomial, namely $a,a^2,a^4$. The Frobenius map $x\mapsto x^2$ realizes those three choices. Hence
$$
N_G(P)=P\rtimes C_3,\qquad |N_G(P)|=21,\qquad n_7(G)=8.
$$
If $7\mid |H|$ for a proper $H<G$, its Sylow count is one or eight. Eight would make $56\mid|H|$, hence $|H|=56$. Its index-three <coset> action would inject the <simple group> $G$ into $S_3$: the kernel is normal and cannot be all of $G$. This is impossible. Thus $H$ normalizes a unique Sylow 7-subgroup and lies in the order-21 <normalizer>.
If $7\nmid |H|$, its order divides 24. In its action on seven nonzero <vectors>, an odd orbit has size one or three. A one-point orbit gives a fixed point. If a three-point orbit is collinear, it is the set of nonzero <vectors> of an invariant plane. If its three <vectors> are independent, their nonzero sum is fixed by $H$. Thus $H$ lies in a point or plane <stabilizer subgroup>.
The index-seven <stabilizer subgroups> are maximal, and any proper overgroup of the order-21 <normalizer> would again have a unique Sylow 7-subgroup and normalize that same <subgroup>, so the <normalizer> is maximal too. The point and plane <stabilizer subgroups> are not conjugate: a <stabilizer subgroup> fixes a point, whereas a plane <stabilizer subgroup> is transitive on its three internal points and its four external points, so fixes no point. Each type is one <conjugacy class>. We have proved <maximal subgroups of GL3 over F2>:
$$
\boxed{GL_3(2):\text{ three classes, of orders }21,\ 24,\ 24.}
$$
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