What real numbers do algebraic numbers cover? Hardy and Wright mention ( though don't give a proof ) that any finite combination of real quadratic surds is an algebraic number. For example
$\sqrt{11+2\sqrt{7}}$. Are all finite combinations of cube root, fourth root ... $n^{th}$ root also algebraic ? such as $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$.

Yes, this holds in general. Recall that an extension $F/\Bbb Q$ is algebraic iff for each $\alpha\in F$ we have

$$[\Bbb Q(\alpha):\Bbb Q]=\dim_{\Bbb Q} \Bbb Q(\alpha)< \infty$$

that is, if every element in it generates a finite extension. Similarly any finite extension is algebraic since if $\beta\in F$ then $\Bbb Q(\beta)\subseteq F\implies [\Bbb Q(\beta):\Bbb Q]\le [F:\Bbb Q]$. For a nested radical expression, this is easy then, take the most nested radical, adjoin the needed roots and proceed by induction.
In your first example $\alpha=\sqrt{11+2\sqrt 7}$, then we see that the field $\Bbb Q(\sqrt 7)$ has a polynomial for this which is ${1\over 2}(x^2-11)-\sqrt 7 = 0$ so the splitting field of this polynomial, which has degree at most $2$ over $\Bbb Q(\sqrt 7)$, and therefore degree at most $4$ over $\Bbb Q$, is algebraic, hence $\alpha$ is algebraic.
For your other example and more generally any example, you do similarly, since there are only finitely many nested radicals each time, you get splitting fields of finite degree, implying all such expressions are algebraic.
Just to work out the more complicated example, $\alpha = \sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$ first adjoin $\sqrt 6$, and let $F_1=\Bbb Q(\sqrt 6)$. Then look at the splitting field, $F_2$, of $(x/3)^7-5-3\sqrt 6$ over $\Bbb Q(\sqrt 6)$. Then the splitting field, $F_3$ of $x^3-2-3\sqrt[7]{5+3\sqrt 6}$ over $F_2$, finally let $F_4=F_3(\sqrt[9]{2})$. Because each step had finite degree, $[F_4:\Bbb Q]<\infty$ showing $\alpha$ is algebraic.
One more since we're on a roll, look at $\sqrt{\sqrt{23}+\sqrt 5}$ this gives you a slight problem at first it seems since $\sqrt{23}$ and $\sqrt{5}$ are both the most nested radical, but as usual $[F(\alpha,\beta): E]\le [F(\alpha):E][F(\beta):E]$ and we can reduce to the simpler case by the classical result that sums and products of algebraic numbers are algebraic (proved in the same way with finite degrees).
