Here's what I've got so far (and I'm assuming $\alpha$ is a unit speed curve):
a) The fact that $\beta(s) = \alpha(s) + r(s)N(s)$ for some scalar function $r$ follows trivially because of the fact that the normal lines of $\beta$ and $\alpha$ are equal (so $\beta(s)$ is a point on the normal line of $\alpha(s)$, which is precisely what the equality states). Then:
$$\ r(s) = (\beta(s) - \alpha(s))N(s) \Rightarrow r'(s) = (\beta'(s) - \alpha'(s))N(s) + (\beta(s) - \alpha(s))N'(s) $$
$$r'(s) = (\beta'(s) -\alpha'(s))N(s) + r(s)N(s)N'(s) = (\beta'(s) - \alpha'(s))N(s)= \beta'(s)N(s) = 0$$
as desired.
b) Let $\theta$ be the angle between the unit tangent vectors mentioned. Then: $$\cos(\theta) = \frac{\beta'(s)\alpha'(s)}{||\beta'(s)||} = \frac{1-r(s)k(s)}{||\beta'(s)||}$$
but $$||\beta'(s)||^2 = ||\alpha'(s)||^2 + r^2(s)||N(s)||^2 = 1 + r^2(s)$$ so $$||\beta'(s)|| = \sqrt{1+r^2(s)}$$ $$\cos(\theta) = \frac{1-r(s)k(s)}{\sqrt{1+r^2(s)}}$$
EDIT: The formula for $||\beta'(s)||^2$ above is wrong. Actually, $||\beta'(s)||^2 = ||\alpha'(s)||^2 + r^2(s)||N'(s)||^2$, which is not as neat as the above. But the hint in the comments is much better to solve this.
c) Assuming I had done b), then the following would be true: $$(\cos(\theta))' = \left( \frac{\beta'(s)\alpha'(s)}{||\beta'(s)||} \right)' = 0 \Rightarrow (\beta'(s)\alpha'(s))' = 0 \Rightarrow \beta''(s)\alpha'(s) + \beta'(s)\alpha''(s) = 0$$
but continuing on this path only leads to $k'(s) = 0$, which, even if I had done b), would not be useful (and I'm aware using this to prove b) is circular).
Beyond these (b) and c)), I would also like some help solving d) and $21$ (I don't know where to start on those).
Update: A lot of what I wrote above is wrong, but my doubts here are almost completely solved. I just have to finish $21$ and I'll be done.
Update 2: For $20$, see the discussion in comments. For $21$:
Since $T_\beta$ is orthogonal to $N_\alpha$, we can write it as a linear combination of $T_\alpha$ and $B_\alpha$. In particular: $$T_\beta = \pm (\cos(\theta)T_\alpha + \sin(\theta)B_\alpha)$$ where $\theta$ is the angle between the unit tangent vectors of $\alpha$ and $\beta$. Then: $$B_\beta = T_\beta \times N_\beta = \pm T_\beta \times N_\alpha = \pm (\cos(\theta)T_\alpha + \sin(\theta)B_\alpha) N_\alpha= \pm(\cos(\theta)B_\alpha - \sin(\theta)T_\alpha)$$ Differentiating and using the fact that $\cos(\theta)$ and $\sin(\theta)$ are constant, we have: $$B_\beta' = \pm(\cos(\theta)B_\alpha' -\sin(\theta) T_\alpha') = \pm(-\cos(\theta)\tau_\alpha N_\alpha - \sin(\theta)\kappa_\alpha N_\alpha)$$ $$B_\beta' = \pm(-\cos(\theta)\tau_\alpha - \sin(\theta)\kappa_\alpha)N_\alpha$$
By the Frenet frame, we also have: $$B_\beta' = -\tau_\beta N_\beta = \pm(-\tau_\beta N_\alpha) $$
So:
$$\pm(-\cos(\theta)\tau_\alpha - \sin(\theta)\kappa_\alpha)N_\alpha = \pm(-\tau_\beta N_\alpha) \Rightarrow (\cos(\theta)\tau_\alpha + \sin(\theta)\kappa_\alpha) = \pm \tau_b $$
By previous work:
$$\cot(\theta) = \frac{1-r(s)\kappa_\alpha(s)}{r(s) \tau_\alpha(s)}$$
which is equivalent to: $$\kappa_\alpha(s) + \cot(\theta) \tau_\alpha = \frac{1}{r(s)} \Rightarrow \kappa_\alpha(s) \sin(\theta) + \tau_\alpha(s) \cos(\theta) = \frac{\sin(\theta)}{r(s)}$$
Then:
$$\pm (\cos(\theta)\tau_\alpha + \sin(\theta)\kappa_\alpha) = \pm \frac{\sin(\theta)}{r(s)} = \tau_\beta(s)$$
$\bar{s}$ being the arclength paramater, we also have:
$$T_\beta = \frac{d \beta}{ds} \frac{ds}{d\bar{s}}$$
so:
$$T_\beta = \left( ( 1 - r(s) \kappa_\alpha(s))T_\alpha + r(s) \tau_\alpha(s) b_\alpha(s) \right ) \frac{ds}{d\bar{s}}$$
comparing the very first expression for the coefficients, we see that: $$\pm \sin(\theta) = r(s) \tau_\alpha(s) \frac{ds}{d \bar{s}}$$
so that:
$$\frac{d \bar{s}}{ds} = \frac{r(s)\tau_\alpha(s)}{\sin(\theta)}$$
and then, by previous work:
$$\pm \frac{\tau_\alpha(s) \tau_\beta(s) r(s)}{\sin(\theta)} = \pm \frac{\sin(\theta)}{r(s)} \Rightarrow \tau_\alpha(s) \tau_\beta(s) = \frac{\sin^2{\theta}}{r^2(s)}$$
as desired.
Asked and answered (see updates). :)