$n$th derivative of $e^{ax}\sin(bx+c)$ 

  
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*How can we substitute $r \cos \alpha$ and $r \sin \alpha$ for $a$ and $b$?
  
*How, on successive differentiation, is there another $r$ multiplied? 
  


First, note that the title in the excerpt is incorrect. It should be "$n$th derivative of $e^{ax}\sin(bx+c)$".
(1) Here we introducing new quantities $r$ and $\alpha$, so we may define them however we want. Essentially, this amounts to writing the pair $(a, b)$ in polar coordinates, as $(r, \alpha)_{\textrm{polar}}$.
More explicitly, any such $r$ satisfies $$a^2 + b^2 = (r \cos \alpha)^2 + (r \sin \alpha)^2 = r^2 (\cos^2 \alpha + \sin^2 \alpha) = r^2 ,$$
so, $$r = \pm \sqrt{a^2 + b^2}.$$
Now the length of $$\left(\frac{a}{r}, \frac{b}{r}\right)$$ is
$$\sqrt{\left(\frac{a}{r}\right)^2 + \left(\frac{b}{r}\right)^2} = \sqrt{\frac{a^2 + b^2}{r^2}} = \sqrt{\frac{r^2}{r^2}} = 1 ,$$
and so it lies on the unit circle. In particular, since $\theta \mapsto (\cos \theta, \sin \theta)$ parameterizes all of unit circle, there is some $\alpha$ such that $$\frac{a}{r} = \cos \alpha, \qquad \frac{b}{r} = \sin \alpha,$$
and rearranging gives the equations in the text, namely,
$$a = r \cos \alpha, \qquad b = r \sin \alpha .$$
Since $(r \cos (\alpha + \pi), r \sin(\alpha + \pi)) = (-r \cos \alpha, -r \sin \alpha)$, by possibly adding $\pi$ to $\alpha$ we may as well assume that $r \geq 0$, that is that $r = \sqrt{a^2 + b^2}$.
(2) The computation for $y_1$ gives that
$$y_1 = r e^{\alpha x} \sin (b x + c + \alpha) .$$
If we want to compute the next derivative, we have
$$y_2 = (r e^{\alpha x} \sin (b x + c + \alpha))' = r (e^{\alpha x} \sin (b x + c + \alpha))$$
But if we replace $c$ with $c + \alpha$ in our rule, it tells us that $(e^{\alpha x} \sin (b x + c + \alpha) = r e^{\alpha x} \sin (b x + c + 2 \alpha)$, and substituting back in the previous display equation gives 
$$y_2 = r (r e^{\alpha x} \sin (b x + (c + \alpha) + \alpha)) = r^2 e^{\alpha x} \sin (b x + c + 2 \alpha)$$
as claimed.
