= Solution
A finite <separable field extension> of degree $d$ has exactly $d$ embeddings into an algebraic closure: adjoin finitely many generators successively, and use part (i) to count extensions of each embedding by the distinct roots of the relevant separable minimal <polynomial>.
For $L=K(\alpha,\beta)$, label these embeddings $\sigma_1,\ldots,\sigma_d$. For each pair $i\ne j$, the equality
$$
\sigma_i(\alpha)+c\sigma_i(\beta)=\sigma_j(\alpha)+c\sigma_j(\beta)
$$
excludes at most one $c\in K$. If the images of $\beta$ agree, the images of $\alpha$ cannot agree too, so it excludes none. Since $K$ is infinite, choose $c$ avoiding the finitely many bad values and put $\eta=\alpha+c\beta$. The $d$ images of $\eta$ are distinct, hence its minimal <polynomial> has degree at least $d$. But $K(\eta)\subset L$ gives degree at most $d$. Therefore $K(\eta)=L$. Combining generators two at a time proves the <primitive element theorem> for every finite separable extension over an infinite base field.
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