Consistent and conservative extensions Let $K_0$ be a consistent $n$-th order theory, for some fixed $n \in \{1, 2, \dots\}$. Suppose $K_1$ is a consistent extension of $K_0$, and suppose $K_2$ is a conservative extension of $K_0$.
Is $K_1 \cup K_2$ necessarily consistent? If so, is $K_1 \cup K_2$ necessarily a conservative extension of $K_1$?
 A: No, $K_1\cup K_2$ is not necessarily consistent, hence neither is it necessarily a conservative extension of $K_1$.
As a counter-example consider the first order formal language $L_0$ with signature consisting of the constant $0$ (zero), the binary function $+$ (addition) and the binary predicate $=$ (equality).
Denote by $K_0$ the $L_0$-theory consisting of the equality axioms, of the (additive) Abelian group axioms, and of the axiom $\exists x, x \neq 0$. Any Abelian group with cardinality $\geq 2$ is a model for $K_0$, hence $K_0$ is consistent.
Denote by $L_1$ the language obtained from $L_0$ by adding to its signature a unary function $-$ (minus).
Denote by $K_1$ the $L_1$-theory obtained by augmenting $K_0$ with the following axiom: $\forall x, -x=x$. Any model for $K_0$ can be transformed into a model for $K_1$ by interpreting $-$ as the identity function. Hence $K_1$ is consistent.
Denote by $K_2$ the $L_1$-theory obtained by augmenting $K_0$ with the following axiom: $\forall x, x+(-x)=0$. $K_2$ is a conservative extension of $K_0$ by the conservativity theorem (for a proof, see [1] Proposition 2.28, pp. 102-103 (link)).
However, the theory $K_1\cup K_2$ is inconsistent.

[1] Elliott Mendelson (2015). Introduction to Mathematical Logic (6th ed.) CRC Press.
