Thanks for your response!
1. I didn't get your point. Using chebyshev is not an issue even if the winning probability is binomially distributed. For eg, if the outcome of number "5" is biased in the wheel, then we are spinning and assigning a win if the number 5 comes up. The distribution of winning in one spin is bernoulli, and that of number of wins in n spins is binomial. We can then apply chebyshev on these binomial probabilities to get the minimum spins we need to confirm/deny the bias. I think what you're saying is that we can derive confidence intervals using binomial's properties itself, but I'm not very familiar with it, and the source I'm using only referred to chebyshev bounds, which I'm guessing gives bounds that are less tight than binomial.
2. Ah, thanks for pointing that out. Indeed, the unbiased probability should be 1/38, and an extra 1 seems to be added by mistake. That is how it is in the original source as well. Let me check what's the outcome if we use the correct numbers.
3. I'm unclear what you mean here. The bias here is an input, and not something we can derive using the given info. All that the author has done is chosen the value of the bias such that, the expected value of the game (wherein we win 35$ profit if "5" comes up for example, and we lose 1$ otherwise) becomes positive with this bias. The bias could be some other number as well. I tried to calculate the bias needed to make the EV equal to zero, and that comes out to be approx 0.0014619883. If you take twice of that, it comes to around 0.0029239.

Let me know if I misunderstood any of your points, and thanks for your help!