I'm reading Courant's Differential and Integral Calculus, there is this session of exercises:
My trouble here is that previously he talks about the rational numbers, why it is needed for them to be extended, he talks a little bit about real numbers and their representation as infinite decimals, expression of numbers in scales other than $10$, inequalities and the Schwarz's Inequality. My question is: Are we supposed to prove all these using only what was given before in the book? We have two classes of exercises:
Exercises in which it is obvious that you need what was given previously: It is quite obvious that to solve - for example - $6,8,etc$ you can use what was given in the sections about inequalities, Schwarz' inequality, etc.
Exercises in which it is not obvious that you need what was given previously: For example, to solve $1.a$, I used the common proof with factorization of integers, $1d$ needs the test of rational roots, which was not given previously unless it was under some disguise, $2$ might need something else but I have been thinking if the Schwarz' inequality could be used to prove it, I have been doing the same (Schwarz) for the other examples without success.
This is important for me because I'm not sure what tools I am allowed to use and I know that there could be some out-of-the-box technique in which it's much easier to prove things, for example: For the examples in $5c$, I could use maxima/minima from single-variable calculus, for $5a,c$ I could use maxima/minima from multivariable calculus, but that would take all the fun from it.

Since it's a calculus book, I would assume that anything precalculus including algebra and trigonometry is fair game to use.
In my opinion this falls under the algebra/precalculus umbrella. Anyway, you can essentially derive the rational root theorem in much the same way you proved that $\sqrt{2}$ is irrational without explicitly using the rational root theorem for $x^2-2=0$.
Remember the $\tan(\alpha-\beta)$ formula, and use that $\tan(\pi / 3)$ is irrational.
Easier: write it as $\displaystyle\,x^2\left(x^2+\frac{1}{x^2}-3\left(x+\frac{1}{x}\right)+4\right)=x^2(y^2-3y+2)\,$ where $\displaystyle\,y=x+\frac{1}{x}\,$.
Easier: note for example that $\displaystyle\,x^2+xy+y^2=\frac{x^3-y^3}{x-y}\,$.