Since $f$ is irreducible, the simple subextension $F(α_1)$ of $K$ is isomorphic to $F[t] /⟨f⟩$, which has dimension $n$ over $F$. Hence $[F(α_1): F]=n$. By the Tower Law, we conclude that $n ∣[K: F]$. But $|G|=[K: F]$ by FTGT.
Another method: $G$ acts transitively on the roots of $f$, so by the Orbit-Stabilizer Theorem, $|\operatorname{Orbit}(α_1)|=n$ divides the order of $G$.

Consider indeterminates $x_1, ⋯, x_n$ and let $V(x):=(x_j^{i-1})_{i, j=1}^n$. For a fixed pair $1 ⩽ i<j ⩽ n$, consider the substitution map $Ψ_{i, j}: ℤ[x_1, ⋯, x_n] → ℤ[x_1, ⋯, x_n]$ which sends $x_j$ to $x_i$ and sends $x_k$ to $x_k$ for all $k ≠ j$. Then $\ker Ψ_{i, j}=⟨x_j-x_i⟩$ and $Ψ_{i, j}(\det V(x))=0$ because two columns are same. Since the $\binom{n}{2}$ linear polynomials $\{x_j-x_i: 1 ⩽ i<j ⩽ n\}$ in $ℤ[x_1, ⋯, x_n]$ are coprime and since this ring is a UFD, we deduce $\prod_{1 ⩽ i<j ⩽ n}(x_j-x_i) ∣ \det V(x)$. Both of these expressions are homogeneous polynomials of degree $1+2+⋯+n-1=\binom{n}{2}$, and the coefficient of $1 ⋅ x_2 ⋅ x_3^2 ⋯ x_n^{n-1}$ in both is equal to 1. Hence $\det V(x)=\prod_{1 ⩽ i<j ⩽ n}(x_j-x_i)$.
The determinant of $f$ is defined to be $δ_f:=\prod_{1 ⩽ i<j ⩽ n}(α_j-α_i)$. The discriminant of $f$ is $Δ_f:=δ_f^2$. Evaluate $x_i$ at $α_i ∈ K$ to see that $δ_f=\det V(α)$ and hence $Δ_f=δ_f^2=(\det V(α))^2$.

Since $f$ is separable, $x_i$ are distinct, $δ_f≠0$.
$S_n$ acts on $ℤ[x_1, ⋯, x_n]$ by permuting the variables. If we swap any two columns of $V(x)$, then the determinant changes sign: $τ ⋅ \det V(x)=\det(τ ⋅ V(x))=-\det V(x)$ for any transposition $τ ∈ S_n$. Since the transpositions generate $S_n$, we see that $σ ⋅ \det V(x)=\operatorname{sgn}(σ) \det V(x)$ for all $σ ∈ S_n$. Since $δ_f=\det V(α)$ by the above, it follows that $σ ⋅ δ_f=\operatorname{sgn}(σ) δ_f$ for all $σ ∈ G$. If $Δ_f$ is a square in $F$, then $δ_f ∈ F$, but $G$ fixes every element of $F$, so $σ ⋅ δ_f=δ_f$ for all $σ ∈ G$ [this argument doesn't work in characteristic 2] Hence $\operatorname{sgn}(σ)=1$ for all $σ ∈ G$ and $G ⩽ A_n$.

Let $x, y, z$ be the three given quantities and let $g(t)=(t-x)(t-y)(t-z)∈F[t]$. First, we show that $Δ_g=Δ_f$ : we have \[ x-y=\left(α_1-α_4\right)\left(α_2-α_3\right),   y-z=\left(α_1-α_2\right)\left(α_3-α_4\right)   \text{ and }   z-x=\left(α_1-α_3\right)\left(α_4-α_2\right) \] which implies that \[ Δ_g=(x-y)^2(y-z)^2(z-x)^2=\left(\left(α_1-α_4\right)\left(α_2-α_3\right)\left(α_1-α_2\right)\left(α_3-α_4\right)\left(α_1-α_3\right)\left(α_4-α_2\right)\right)^2=Δ_f \] Hence $Δ_g=Δ_f$ as claimed. Since $x, y, z$ are assumed to lie in $F$, it follows that $Δ_g$ is a square in $F$. Hence $Δ_f$ is a square in $F$ as well, so $G ⩽ A_4$ by part (c). On the other hand, $4|{|G∣}$ by part (a) because $f$ is irreducible. Hence $G=V_4$ or $G=A_4$.
Suppose for a contradiction that $G=A_4$. Then $G$ contains the three-cycle (123). However $(123) ⋅ x=α_2 α_3+α_1 α_4=z$. Since $x ∈ F ⊆ K^G$, we then get $z=(123) ⋅ x=x$. But this implies that $z-x=\left(α_1-α_3\right)\left(α_2-α_4\right)=0$, so either $α_1=α_3$ or $α_2=α_4$, contradicting the separability of $f$.