= Solution
Write $q=m/s$, an integer under the stated divisibility assumption. The degree relation gives $n=rm/s=qr$. Take the <disjoint union of graphs> of $q$ copies of the <complete bipartite graph> $K_{s,r}$, placing its $s$-vertex part in $U$ and its $r$-vertex part in $W$. The total part sizes are then $qs=m$ and $qr=n$. Each left <vertex> has <vertex degree> $r$, and each right <vertex> has <vertex degree> $s$, so the graph has exactly the required <biregular graph> parameters.
An <independent set> in one component can be any subset of its left part or any subset of its right part. It cannot contain a vertex from both parts, because all cross-pairs are <edges>. The two families share only the empty set, so the component has $2^s+2^r-1$ <independent sets>. Choices in different <connected components of a graph> are independent combinatorial choices, and their counts multiply. Hence
$$
\boxed{i(G)=(2^s+2^r-1)^q=(2^r+2^s-1)^{m/s}.}
$$
This attains equality in part (i). It also explains the entropy weight $2^r$ there: the empty selected left neighbourhood permits all $2^r$ subsets of the opposite part.
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