$6\times 6$ = 36, a perfect square, hence $\sqrt{36}$ is not irrational .
Also 25 is the perfect square of 5, hence $\sqrt{25}$ is rational.

$4^2=16$ & $9^2=81$, hence all the numbers between  $\sqrt{16}$ and $\sqrt{81}$  lie between 4 and 9.

We can see that  $\sqrt{17}$, $\sqrt{31}$   are irrational numbers  between $\sqrt{16}$ and $\sqrt{81}$, and they lie between 4 and 9.

$2\sqrt{3}\times 3\sqrt{2}=2\times 3\times \sqrt{2\times 3}=6\sqrt{6}$
The presence of factor $\sqrt{6}$ makes the expression irrational.
 

Given:
${\left(\sqrt{1024}\right)}^3=8^x$

$\Rightarrow \left(\sqrt{(}{2}^{10}\right){)}^3=8^x$.   
$\Rightarrow \left(\sqrt{(}\right(2^5{)}^2){)}^3=8^x$.  {as ${\left(a^m\right)}^n$=$a^{mn}$}

${\Rightarrow \left(2^5\right)}^3={\left(2^3\right)}^x$

${\Rightarrow \left(2^3\right)}^5={\left(2^3\right)}^x$
$\Rightarrow (8{)}^5$= $(8{)}^x$
We know that, if $a^n$= $a^m$ then, $n=m$.
Hence,
$\Rightarrow x=5$

${\left(\frac{7}{9}\right)}^3=\frac{7\times 7\times 7}{9\times 9\times 7}\Rightarrow {\left(\frac{7}{9}\right)}^3=\frac{343}{729}$

Rationalizing the expression we get,
$\frac{1}{\sqrt{5}-\sqrt{7}}=\frac{\sqrt{5}+\sqrt{7}}{(\sqrt{5}-\sqrt{7})(\sqrt{5}+\sqrt{7)}}$ 

=$\frac{\sqrt{5}+\sqrt{7}}{(\sqrt{5}{)}^2-(\sqrt{7}{)}^2}$

=$\frac{\sqrt{5}+\sqrt{7}}{5-7}$
=$\frac{-(\sqrt{5}+\sqrt{7})}{2}$

We know that $\sqrt[n]{na}\times \sqrt[n]{nb}=\sqrt[n]{nab}$
$\text{So,}\sqrt[n]{n2}\times \sqrt[n]{n3}=\sqrt[n]{n2\times 3}=\sqrt[n]{n6}$