\(\displaystyle \sqrt[3]{7+5\sqrt{2}} = x+y\sqrt{2}\)

Cube both sides:

\(\displaystyle 7+5\sqrt{2} = \left(x+y\sqrt{2}\right)^3 = x^3+3x^2y\sqrt{2}+6xy^2+2y^3\sqrt{2} = (x^3+6xy^2) + (3x^2y+2y^3)\sqrt{2}\)

Equating coefficients, we find \(\displaystyle x^3+6xy^2 = 7\) and \(\displaystyle 3x^2y+2y^3 = 5\). Looking quickly, I realize that if \(\displaystyle x=y=1\), the system of equations is solved. Hence, \(\displaystyle \sqrt[3]{7+5\sqrt{2}} = 1+\sqrt{2}\)

For the next one, do the same thing:

\(\displaystyle \sqrt{\dfrac{11}{2}-3\sqrt{2}} = x+y\sqrt{2}\)

Square both sides:

\(\displaystyle \dfrac{11}{2}-3\sqrt{2} = \left(x+y\sqrt{2}\right)^2 = x^2+2xy\sqrt{2}+2y^2\)

Equating coefficients, we have \(\displaystyle x^2+2y^2 = \dfrac{11}{2}\) and \(\displaystyle 2xy = -3\)

Let's solve for \(\displaystyle y\) in the equation on the right: \(\displaystyle y = -\dfrac{3}{2x}\) (we know \(\displaystyle x \neq 0\) because otherwise, \(\displaystyle 2xy=0\neq -3\)).

Plugging that into the first equation:

\(\displaystyle \begin{align*}x^2+2\left(-\dfrac{3}{2x}\right)^2 & = \dfrac{11}{2} \\ x^2 + \dfrac{9}{2x^2} & = \dfrac{11}{2} \\ 2x^4+9 & = 11x^2 \\ 2x^4-11x^2+9 & = 0 \\ (2x^2-9)(x^2-1) & = 0\end{align*}\)

Hence, \(\displaystyle 2x^2-9=0\) or \(\displaystyle x^2-1 = 0\). Since \(\displaystyle x\) must be a rational number, it cannot be the first one, so \(\displaystyle x^2-1=0\). Then \(\displaystyle x = \pm 1\).

Before, we found \(\displaystyle y = -\dfrac{3}{2x}\). So, if \(\displaystyle x=1\), then \(\displaystyle y = -\dfrac{3}{2}\). If \(\displaystyle x=-1\), then \(\displaystyle y = \dfrac{3}{2}\).

So, \(\displaystyle \sqrt{\dfrac{11}{2}-3\sqrt{2}} = \pm \left(1-\dfrac{3}{2}\sqrt{2}\right)\)

Checking on a calculator, you find it is \(\displaystyle \sqrt{\dfrac{11}{2}-3\sqrt{2}} = -1+\dfrac{3}{2}\sqrt{2}\).