The first is true in any hereditarily normal space: separated sets have disjoint neighbourhoods. It fails in the compact product $(\omega_1+1)\times(\omega+1)$ (Tychonoff's plank with corner point). The set $C=\{(\alpha,\beta): \alpha=\omega_1$ or $\beta=\omega\}$ is closed. The sets $U=\{(\alpha,\omega):\alpha<\omega_1\}$ and $V=\{(\omega_1,n):n<\omega\}$ are open-in-$C$ but have no disjoint extensions.
Similarly, in the second statement $A$ and $B$ are already closed-in-$X$, so the second statement is true for hereditarily normal spaces and false for the same example.
Addendum: the first statement characterizes hereditary normality: if $A$ and $B$ are separated let $C=\overline{A\cup B}$ and $U=C\setminus\overline{B}$ and $V=C\setminus\overline{A}$. Then $U$ and $V$ are open in $C$, with $A\subseteq U$ and $B\subseteq V$. Then $U'$ and $V'$ would be disjoint neighbourhoods of $A$ and $V$ respectively.