= Solution
A <regular category> has <finite limits>, factorizations of every arrow as a <regular epimorphism> followed by a <monomorphism>, and pullback-stable <regular epimorphisms>. Equivalently one may require <coequalizers> of <kernel pairs> and pullback stability of the resulting quotients. A <regular epimorphism> is a <coequalizer> of some parallel pair; an <extremal epimorphism> is an <epimorphism> $e$ such that $e=mh$, with $m$ monic, forces $m$ to be invertible.
First, every <regular epimorphism> is extremal without assuming regularity of the whole category. Suppose $e:A\to B$ coequalizes $r,s$ and $e=mh$. Since $m$ is monic, $hr=hs$, so the universal property gives $k:B\to\operatorname{dom}(m)$ with $ke=h$. Then $mke=e$; cancellation of the <epimorphism> $e$ gives $mk=1_B$. Monicity of $m$ now gives $km=1$, so $m$ is an <isomorphism>. Conversely, in a <regular category>, factor an extremal $e$ as $mq$ with $q$ regular epic and $m$ monic. Extremality makes $m$ invertible, so $e$ is regular epic. Thus \b[covers and <regular epimorphisms> coincide].
For a counterexample, work in the category of small categories, which has <finite limits> constructed on objects and arrows. Let $C$ be two disjoint walking arrows $a\xrightarrow{u}b$ and $c\xrightarrow{v}d$. Let $D$ be the walking <isomorphism>, with two objects $0,1$ and inverse arrows $p:0\to1$, $q:1\to0$. Define $F:C\to D$ by $a,d\mapsto0$, $b,c\mapsto1$, $u\mapsto p$, $v\mapsto q$. These images generate all of $D$, so two functors agreeing after $F$ agree everywhere; thus $F$ is epic.
A monic functor is injective on objects and faithful, as tested by functors from the terminal category and from the walking arrow. If $F$ factors through such a functor $m:D'\to D$, its image contains both objects and both arrows $p,q$. Faithfulness forces their lifted composites to be the identities, so the image contains every arrow of $D$ and $m$ is an <isomorphism>. Hence $F$ is extremal.
It is not regular. Its <kernel pair> identifies the endpoint pairs $a,d$ and $b,c$ but has no relation forcing the composites of the images of $u,v$ to be identities. Concretely, send all four objects to the one-object category with endomorphism monoid $\{1,e\}$, $e^2=e\ne1$, and send both $u,v$ to $e$. This functor equalizes the <kernel pair>: corresponding identities agree, and the only nonidentity arrow pairs in the <kernel pair> are an arrow paired with itself. It cannot factor through $D$, because images of $p,q$ would have to be inverse, whereas $e^2\ne1$. In a category with <kernel pairs>, every <regular epimorphism> is the <coequalizer> of its <kernel pair>: its original coequalized pair factors through that <kernel pair>, giving the universal property. Therefore \b[$F$ is an extremal but nonregular <epimorphism>], the promised <nonregular extremal epimorphism in the category of categories>.
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