The elevator has to perform more work in the moment the person is accelerating upwards, but then again less work when stopping at the top, i.e. decelerating. In total, the work performed by the elevator is therefore the same as if the person was standing still the whole time.
Yes, when accounting for factors such as air resistance and biological processes, there is clearly energy used, and work being done somewhere.
If you're only considering a book moving at a constant velocity in the horizontal direction, you could say there is no net work done on the book.
If we analyze it further, you will find that if it moves with a constant velocity, there must be some other force acting on it to counteract the drag force. This is the force supplied by you when moving the book.
If this book is moving with constant velocity, this must mean that the net force acting on the book in the horizontal direction is $0$. If that is the case, the net work in the horizontal direction required to keep it moving at the horizontal velocity is zero. The work done by the person on the book is exactly countered by the work done by the air resistance on the book, and thus no net work.
Work is required for a person to move a book through the air at a constant velocity, because the work done by air resistance needs to be countered, but the net work on the book during that movement is none.
It's also often the case that it's assumed you can neglect air resistance. In that case, it would not require work to keep the book moving at a constant velocity. I suspect they may be making that assumption when they are discussing work.
You could also consider a situation where you are starting from a standstill, and ending at a standstill while covering some distance. You can see there is work done to accelerate and decelerate the book. From an energy/work perspective, the work required to slow down the book is exactly opposite of the work required to get it up to the moving speed. This also means that no net work is done on the book during the entire travel (though both the person and the air will have had net work done on them).
Best Answer
Yes you are right, $W=f.d$, but this is a scalar product and $f$ and $d$ are vectors. Which means that the norm is $W=|f||d|\cos(\alpha)$ where $\alpha$ is the angle of the stairs. One can notice that $h=|d|\cos(\alpha)$ is the height of the stairs.
In other words, if you project all the vectors on the vertical axis, it becomes $W=|f||d|\cos(\alpha)=|f|h$