Evaluating $\int_0^1 \frac{3x}{\sqrt{4-3x}} dx$

This is very convoluted solution. It would be easier if you make a substitution $t= 4-3x$, then you get $$\int {4-t\over -3\sqrt{t}}dt =-{1\over 3}\int (4t^{-1/2}-t^{1/2})dt=... $$

Your second question is indeed, despite your doubts, about the substitution method. The reason you don't see a $du$ (or any other variable change) is because they did it in a less-common way notationally, but if you try it with substitution you will see that it all works out the same.

The first question is a good one, because it is often not enough to know that something works, but instead you need to know how to do similar things in the future. The solution writer noticed that the integral was difficult as written, but that they could use substitution (as mentioned above) if they had two rational expressions, each containing $x$ in just one place. Since the $4$ was under the square root, they went ahead and tried adding and substracting that from the numerator to get the square of the denominator, cancelling it, on one rational, and just a constant on the other. It worked in this case. It won't always work.

Try the same logic with the integrands below to see how it goes:

$$\dfrac{2x}{5-2x}$$

$$\dfrac{2x}{5-3x}$$

Perhaps after trying both of those you can even conjecture about when the method will work and when it won't!

Addressing question 1: \begin{align*} & - \left( \frac {4-3x-4}{\sqrt{4-3x}}\right) \\ = & - \left( \frac {4-3x}{\sqrt{4-3x}} - \frac{4}{\sqrt{4 - 3x}}\right) \\ = & - \frac {(\sqrt{4-3x})^2}{\sqrt{4-3x}} - \left(- \frac{4}{\sqrt{4 - 3x}}\right) \\ = & - \sqrt {4-3x} + \frac {4}{\sqrt{4-3x}} \end{align*}