= Solution
Choose a minimal nontrivial <normal subgroup> $L$ of $K$ itself. Every conjugate $L^g$ for $g\in G$ is also a <minimal normal subgroup> of $K$, since $K$ is normal in $G$. Distinct such subgroups intersect trivially: their intersection is normal in $K$ and minimality forces either equality or trivial intersection. Moreover $[L_i,L_j]\subseteq L_i\cap L_j=1$, so distinct ones commute.
Choose a maximal family of these conjugates whose product $M=L_1\times\cdots\times L_r$ is direct. For any further conjugate $L^g$, its intersection with $M$ is normal in $K$ and is either all of $L^g$ or trivial. In the latter case it commutes with all the existing factors and could be adjoined to the <direct product of groups>, contrary to maximality. Thus every conjugate is contained in $M$. The product of all conjugates is consequently $M$, which is nontrivial and normal in $G$. Minimality of $K$ as a $G$-normal subgroup gives $M=K$.
Any <normal subgroup> of a factor $L_i$ is normal in $K$: the other direct factors commute with it, and $L_i$ normalizes it. Minimality of $L_i$ as a $K$-normal subgroup makes it simple. All factors are isomorphic because they are conjugate to $L$. This proves the <direct-product structure of a finite minimal normal subgroup>:
$$
\boxed{K\cong S^r\quad\text{for a finite simple group }S.}
$$
An abelian <simple group> is cyclic of prime order, so the abelian case is elementary abelian. Now let $H$ be a maximal proper subgroup of a nonabelian <simple group> $X$. It is nontrivial, since $X$ has nontrivial proper <cyclic subgroups>. Choose a <minimal normal subgroup> $K$ of $H$. Then $H\le N_X(K)$. This <normalizer> cannot be $X$, since that would make $1<K\le H<X$ normal in $X$. Maximality gives
$$
\boxed{H=N_X(K),\qquad K\cong S^r.}
$$
For each of the two groups in this question, the allowed absence of proper nonabelian simple subgroups makes the factors of such a $K$ cyclic of prime order. Hence it suffices to examine <normalizers> of nontrivial elementary abelian subgroups.
In $A_5$, the possible primes are two, three and five. The <Sylow subgroups> for the last two primes are cyclic of orders three and five. Every involution is a double <transposition>; for $(12)(34)$ the <centralizer> in $A_5$ is
$$
\{1,(12)(34),(13)(24),(14)(23)\},
$$
the <Klein four-group> fixing point five. Thus an elementary abelian $2$-subgroup is cyclic of order two or a <Klein four-group>. The <normalizer> of a cyclic order-two subgroup equals its <centralizer> and has order four, so is contained in a point <stabilizer> of order twelve. A <Klein four-group>'s common fixed point is intrinsic, and the full point <stabilizer> $A_4$ normalizes it; its <normalizer> is therefore exactly that <stabilizer>.
An order-three subgroup has support a triple. Its <normalizer> preserves the triple and the remaining pair, and is $(S_3\times S_2)\cap A_5$, of order six. An order-five subgroup acts regularly on all five points. Label the points by $\mathbb F_5$, with the subgroup acting by translations. Its <normalizer> in $S_5$ is the twenty affine maps $x\mapsto ax+b$. The translations are even, whereas multiplication by a generator of $\mathbb F_5^{\times}$ is a four-cycle and is odd. Exactly half these maps are even, so the <normalizer> in $A_5$ has order ten.
The only candidates for <maximal subgroups> are thus the <normalizers> of orders twelve, ten and six. They are all maximal: none can lie in a larger candidate, except potentially an order-six subgroup in an order-twelve $A_4$. But an index-two subgroup of $A_4$ would contain every order-three element; there are eight such elements, plus the identity, already more than six. Hence that containment is impossible. Point <stabilizers> are conjugate; order-three and order-five subgroup <normalizers> each form one class by the <Sylow theorems>. Therefore the <maximal subgroups of A5> form precisely
$$
\boxed{\text{three conjugacy classes, of orders }12,10,6.}
$$
Their numbers are five, six and ten respectively, since these proper <maximal subgroups> are self-normalizing in the <simple group>.
For $X=GL_3(2)$, choosing independent columns gives $|X|=(8-1)(8-2)(8-4)=168$. If $t$ is an involution, write $t=I+N$. <Field characteristic> two gives $N^2=0$, and $\operatorname{im}N\subseteq\ker N$ in dimension three forces its nonzero rank to be one. Thus $t=I+vf$, with nonzero vector $v$, nonzero <linear functional> $f$ and $f(v)=0$. This <transvection> fixes the plane $\ker f$ and has image line $\langle v\rangle$. Every flag consisting of an incident line and plane occurs, giving $7\cdot3=21$ involutions, all conjugate. Its <centralizer>, and hence the <normalizer> of its <cyclic subgroup>, is the flag <stabilizer> of order $168/21=8$. It lies in both the line and plane <stabilizers> of order $24$, so cannot be maximal.
For the <elementary abelian subgroups in GL3 over F2>, consider commuting <transvections> $I+vf$, $I+wh$. Their rank-one parts satisfy
$$
v f(w)h=w h(v)f.
$$
If $v,w$ are independent, equality forces both sides zero; thus $f,h$ vanish on $\langle v,w\rangle$ and are the same nonzero functional. If $v=w$, they have a common image line. Consequently the two types of order-four elementary abelian subgroup are
$$
K_L=\{I+vf:f(v)=0\},\qquad
K_W=\{I+wf:w\in W=\ker f\}.
$$
No extra <transvection> can commute with all elements of $K_L$: if its image were different, its functional would have to equal each of the two independent functionals occurring in $K_L$. Dually, the same holds for $K_W$. Thus these exhaust the elementary abelian $2$-subgroups of rank at least two. The <normalizer> of $K_L$ is the line <stabilizer> and the <normalizer> of $K_W$ is the plane <stabilizer>, each of order $168/7=24$. They give different <conjugacy classes> because their common fixed spaces have dimensions one and two. All line <stabilizers> are conjugate, as are all plane <stabilizers>.
For an order-three element, its <minimal polynomial> is $(X-1)(X^2+X+1)$: it has a fixed line and an invariant complementary irreducible plane. Its <normalizer> preserves both and is the full $GL_2(2)$ on that plane, of order six, since the order-three subgroup is normal in $GL_2(2)$. This <normalizer> lies in the corresponding line and plane <stabilizers>, so is not maximal.
An order-seven element is irreducible. Over the <finite field> $\mathbb F_2$ one has $X^7-1=(X-1)(X^3+X+1)(X^3+X^2+1)$, and both cubic factors are irreducible because neither has a root in $\mathbb F_2$. Its <minimal polynomial> divides this square-free product and cannot be just $X-1$, since the element is not the identity. It therefore contains a cubic factor; the three-dimensional space forces that irreducible cubic to be the whole <minimal polynomial>. Identify the space with $\mathbb F_8$ and the element with multiplication by a generator $\alpha$ of $\mathbb F_8^{\times}$. Its <centralizer> consists of the seven nonzero multiplication maps. A <normalizer> induces an $\mathbb F_2$-automorphism of this field, so it sends $\alpha$ to $\alpha$, $\alpha^2$ or $\alpha^4$. All three possibilities occur via the field <automorphisms> $x\mapsto x^{2^j}$. Therefore the <normalizer> of the <Singer cycle> has order $7\cdot3=21$, and all such <normalizers> are conjugate by the <Sylow theorems>.
The maximal-normalizer reduction leaves exactly the two order-twenty-four <stabilizer> types and the order-twenty-one type. Each is maximal because it cannot be contained in any larger remaining candidate; equal-order subgroups cannot properly contain one another and $21$ does not divide $24$. Thus
$$
\boxed{GL_3(2)\text{ has exactly three maximal-subgroup classes, of orders }24,24,21.}
$$
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