What real numbers do algebraic numbers cover?

The usual ways to prove this is to use some basic linear algebra.

Assume $\alpha$ and $\beta$ are algebraic numbers.

The field $\mathbb Q[\alpha][\beta]$ is a finite-dimensional vector space over $\mathbb Q$. Let $d$ be the dimension of the vector space.

Then $1,\alpha+\beta,(\alpha+\beta)^2,\cdots,(\alpha+\beta)^d$ are $d+1$ elements of that vector space, and thus must be linearly dependent. But that means that $\alpha+\beta$ is the root of a rational polynomial of degree at most $d$.

The same for $\alpha\beta$.

More specifically, if $\alpha$ is the root of a rational polynomial of degree $d_1$ and $\beta$ is the root of a rational polynomial of degree $d_2$ then $\alpha+\beta$ and $\alpha\beta$ are both roots of a polynomial of degree at most $d_1d_2$.

Finally, if $\alpha$ is the root of the rational polynomial $p(x)$ then $\sqrt[n]\alpha$ is a root of $p(x^n)$.

So this means that $5+3\sqrt6$ is the root of a polynomial of degree $2$, and $3\sqrt[7]{5+3\sqrt6}$ is a root of a polynomial of degree $14$ and $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt6}}$ is the root of a polynomial of degree $42$ and finally $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$ is a root of a polynomial of degree $42\cdot 9=378$. There might be a polynomial of smaller degree, but we get this upper bound.

Hint: Let $x = \sqrt{11+2\sqrt{7}}$, square this expression $x^2=11+2\sqrt{7} \implies x^2-11=2\sqrt{7}$. Now square again to get a polynomial with rational coefficients. You can use a similar approach for the second expression.

Yes, you can unpack them into a polynomial that they satisfy. For your example, we can write $$x=\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}\\x^3=2+3\sqrt[7]{5+3\sqrt{6}}\\\frac 13(x^3-2)=\sqrt[7]{5+3\sqrt{6}}\\\left(\frac 13(x^3-2)\right)^7=5+3\sqrt 6\\\frac 19\left(\left(\frac 13(x^3-2)\right)^7-5\right)^2-6=0$$

Note: this uses a previous example in the question. The approach is the same. Note that the algebraics are closed under the field operations, so the sum of algebraics is again algebraic.