Solution:

(i) Suppose $A, B, C \in S$ such that $(A, B) \in R$ and $(B, C) \in R$
$\Rightarrow \mathrm{A}$ is a proper subset of $\mathrm{B}$ and $\mathrm{B}$ is a proper subset of $\mathrm{C}$
Therefore, A is a proper subset of $\mathrm{C}$
$\Rightarrow(\mathrm{A}, \mathrm{C}) \in \mathrm{R}$
As a result, $\mathrm{R}$ is transitive.
(ii) Suppose $\mathrm{R}=\{(\mathrm{A}, \mathrm{B}): \mathrm{A} \subset \mathrm{B})\}$, i.e., $\mathrm{A}$ is a proper subset of $\mathrm{B}$ So now,
Any set is a subset of itself, but not a proper subset. $\Rightarrow(\mathrm{A}, \mathrm{A}) \notin \mathrm{R}$
As a result, $R$ is not reflexive.