= Solution
In characteristic two, let $K=\mathbb F_2(t)$ and $L=K(u)$ with $u^2+u=t$. The <Artin–Schreier polynomial> is irreducible: if $r\in K$ solved it, a finite pole of $r$ would create a pole of even order in $r^2+r$, impossible for $t$; at infinity a pole of $r$ likewise gives even order, while $t$ has order one. A constant $r$ cannot solve it either. Its derivative is one, so this degree-two extension is <separable field extension>. Adjoining $\sqrt\alpha$ in characteristic two is either trivial or <purely inseparable field extension>, since the derivative of $X^2-\alpha$ vanishes. Hence $\boxed{L\ne K(\sqrt\alpha)\text{ for every }\alpha\in K}$.
For the second example take $\boxed{K=\mathbb F_{11}}$. Every extension still has characteristic eleven, and $X^{11}-1=(X-1)^{11}$ in it. Thus the only eleventh <root of unity> is one, whose order is one, and there is no primitive eleventh root in any extension.
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