An algebra whose equational theory is finitely based but whose quasiequational theory is not, and vice versa. Is there an algebraic structure $K$ such that its equational theory has a finite basis, but that its quasiequational theory does not have a finite basis? Also, what about vice versa? That is, is there an algebraic structure $K$ such that its equational theory does not have a finite basis, but its quasiequational theory does have a finite basis?
 A: Any nilpotent semigroup is finitely based. (If it is $k$-step nilpotent,
then it satisfies the associative law and the $k$-step nilpotent law. Any further law is equivalent to one of the form $w_1(x_1,\ldots,x_k)\approx w_2(y_1,\ldots,y_k)$ in the fixed set of variables $x_1,y_1,\ldots,x_k, y_k$, and there are only finitely many inequivalent laws in these variables relative to the laws defining the variety of $k$-step nilpotent semigroups.)
On the other hand, a finite nilpotent semigroup with a finitely axiomatizable quasiequational theory  must be a zero (or null) semigroup according to Corollary 1.4 of

RELATIVELY INHERENTLY NONFINITELY
Q-BASED SEMIGROUPS
MARCEL JACKSON AND MIKHAIL VOLKOV
TRANSACTIONS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 361, Number 4, April 2009, 2181-2206
(A null semigroup is one satisfying $x_1x_2\approx y_1y_2$.)
So every finite, non-null, nilpotent semigroup has a finitely axiomatizable equational theory and a nonfinitely axiomatizable quasiequational theory.
For the other part of this question, in Theorem 7.10 of 
FINITE BASIS THEOREMS FOR RELATIVELY CONGRUENCE-DISTRIBUTIVE
QUASIVARIETIES
DON PIGOZZI
TRANSACTIONS
OF THE
AMERICAN MATHEMATICAL
Volume
310, Number
2, December
1988, 499-533
one finds a procedure for starting with any finite algebra $A$
whose equational theory is not finitely axiomatizable, and constructing a new finite algebra, Et$(A)$, whose equational theory is also not finitely axiomatizable, but whose quasiequational theory is guaranteed to be finitely axiomatizable.
If $A$ is Murskii's groupoid or Lyndon's groupoid, then Et$(A)$ provides
the necessary example for the second part.
