The first is true in any [hereditarily normal space][1]: separated sets have disjoint neighbourhoods. 
It fails in the compact product $(\omega_1+1)\times(\omega+1)$ ([Tychonoff's plank with corner point][2]). 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.


  [1]: https://en.wikipedia.org/wiki/Normal_space#hereditarily_normal_space
  [2]: https://en.wikipedia.org/wiki/Tychonoff_plank