Let $w=\frac{1+(1-8\alpha)z}{1-z},\qquad |z|=1,\, z\neq 1$ and we want all real $\alpha$ such that $w$ is purely imaginary for every such $z$.

We must find when $\Re(w)=0$ for all $z$ on the unit circle except $z=1$.

1. Use the standard substitution for points on the unit circle

If $|z|=1$ and $z\neq 1$, write $z=e^{i\theta}, \qquad \theta\in \mathbb R,\ \theta\neq 2k\pi.$

Then $\frac{1+z}{1-z}$ is purely imaginary. Let us verify this carefully.

Using $z=e^{i\theta},$ we get $1+z=1+e^{i\theta}=2e^{i\theta/2}\cos\frac{\theta}{2},$ $1-z=1-e^{i\theta}=-2ie^{i\theta/2}\sin\frac{\theta}{2}.$ So, $\frac{1+z}{1-z}=\frac{2e^{i\theta/2}\cos(\theta/2)}{-2ie^{i\theta/2}\sin(\theta/2)}=i\cot\frac{\theta}{2},$ which is purely imaginary.

2. Rewrite the given expression

We have $w=\frac{1+(1-8\alpha)z}{1-z}.$ Rewrite the numerator: $1+(1-8\alpha)z=(1-z)+2z-8\alpha z$ which is not especially convenient.

A better way is to split it as $1+(1-8\alpha)z=(1+z)-8\alpha z.$ Thus, $w=\frac{1+z}{1-z}-8\alpha\frac{z}{1-z}.$

Now simplify $\dfrac{z}{1-z}$ using $z=(z-1)+1,$ so $\frac{z}{1-z}=\frac{(z-1)+1}{1-z}=\frac{z-1}{1-z}+\frac{1}{1-z}=-1+\frac{1}{1-z}.$ But an even cleaner identity comes from $\frac{1+z}{1-z}=\frac{1-z+2z}{1-z}=1+\frac{2z}{1-z},$ so $\frac{z}{1-z}=\frac{1}{2}\left(\frac{1+z}{1-z}-1\right).$ Therefore, $w=\frac{1+z}{1-z}-8\alpha\cdot \frac12\left(\frac{1+z}{1-z}-1\right).$ Hence, $w=\left(1-4\alpha\right)\frac{1+z}{1-z}+4\alpha.$

3. Analyze real and imaginary parts

From Step 1, $\frac{1+z}{1-z}$ is purely imaginary for all $|z|=1, z\neq 1$.

So in $w=\left(1-4\alpha\right)\frac{1+z}{1-z}+4\alpha,$

• the first term is purely imaginary (because $1-4\alpha\in\mathbb R$),
• the second term $4\alpha$ is purely real.

Therefore, $\Re(w)=4\alpha.$ For $w$ to be purely imaginary for all allowed $z$, we must have $4\alpha=0 \implies \alpha=0.$

4. Check the value

If $\alpha=0$, then $w=\frac{1+z}{1-z},$ which we already proved is purely imaginary for all $|z|=1, z\neq 1$.

So the required set is $\{0\}.$

5. Match with options

Option B is $\{0\}$.

So the correct answer is: $\boxed{\text{B}}$