Label each of $2^n$ rooms by distinct number from $S=\{ 0,1,2,...,2^n-1\}$. For each room number $d\in S$, consider its binary representation, which contains at most $n$ digits. For each $i\in \{ 0,1,2,...,n\}$, let $A_i$, and $B_i$, denote the set of all $d\in S$ that the $i^{th}$ digit (say, from the left) is zero, and one, respectively. Then, in $2n$ moves, send workers to rooms whose number from the set $A_i$ or $B_i$ where $i\in \{ 1,2,...,n\}$. Not hard to check that, for each $d\in S$ and for all $i\in \{ 1,2,...,n\}$, we can determine the $i^{th}$ digit of the number of room whose switch control the bulb in room number $d$ using information we know from sending $A_i$ and $B_i$.

The answer is yes. We use the same method in a) to determine, in first $2n-2$ moves, for each $d\in S$, the $i^{th}$ digit of the number of room whose switch control the bulb in room number $d$ but only for all $i\in \{ 1,2,...,n\}$ but one choice. At this stage, we already get that, for each one of $2^{n-1}$ choices of the determined $n-1$ positions, there're exactly two distinct rooms that the switches in those two rooms control the light bulbs in the two rooms whose the room number appear that fixed $n-1$ positions. So, our objective for the last moves is so that, for every one of $2^{n-1}$ choices of the $n-1$ positions, the following conditions are fulfilled: Exactly one of two rooms whose number contain those $n-1$ positions was chosen, and Exactly one of two rooms whose switch control light bulbs in two rooms in the above condition was chosen. Now this is obvious, view these two conditions as two perfect matching of graph with $2^n$ vertices. Those two perfect matching combined (multiple edge allowed) gives us a graph $G$ such that every vertex has degree $2$. So, $G$ is nothing but union of disjoint cycles. Moreover, all of those cycles are even cycle (why?) For every edge of $G$ connect two vertices $v,w$, we want exactly one of them to be chosen. This can be done easily by just alternatively select and discard vertices along those even cycles, done!

But anyway, it made me search for the original Russian text, here is the beginning: "Euro construction repairs(евроремонт) took place in a building with $2^n$ rooms,...". That term is commonly used in Russia for construction repairs complying with EU standards and using imported European materials. Of course it's much more expensive. So, after such an repair, all things are messed up. I'm far from being paranoid, it's a kind of a joke, but nevertheless it shows some attitude towards EU standards/materials/style of life, etc.