= Solution
An algebraic <field extension> $E/K$ is separable if every $\alpha\in E$ has a separable <minimal polynomial> over $K$.
For an inseparable example, take $K=\mathbb F_p(t)$ and $f(X)=X^p-t$. <Eisenstein's criterion> at the prime $t$ in $\mathbb F_p[t]$, followed by <Gauss lemma for polynomials>, proves irreducibility over its fraction <field>. Its <derivative> is zero, and in a <splitting field> it is $(X-t^{1/p})^p$, so it is inseparable.
Now let $M/K$ be a separable splitting extension of a nonzero <polynomial> $f$. Every irreducible factor $g$ of $f$ has a root $\alpha\in M$, and $g$ is a scalar multiple of the <minimal polynomial> of $\alpha$. Separability of $M/K$ makes this <polynomial> separable. Therefore every irreducible factor of $f$ is separable, which proves the requested conclusion in the factorwise convention. The squarefree qualification under the alternative convention is explained in part (a).
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