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Solution:

Part a

If $\sqrt{4+2\sqrt 3}-\sqrt 3$ is rational. Let us assume it to be $x$, we would like to determine $x$. Then we must have
$$
\sqrt{4+2\sqrt 3}=x+\sqrt{3}.
$$ Taking square of both sides, we obtain
\begin{equation}\label{2-7}
4+2\sqrt{3}=x^2+2x\sqrt 3+3.
\end{equation} Since $x$ is rational, we should expect that $2\sqrt{3}=2x\sqrt 3$ which is true only when $x=1$. Moreover, \eqref{2-7} is true if $x=1$. We have
$$
4+2\sqrt{3}=1+2\sqrt 3+3=(1+\sqrt 3)^2.
$$
Hence $\sqrt{4+2\sqrt 3}=1+\sqrt 3$ and $\sqrt{4+2\sqrt 3}-\sqrt 3=1$ is rational.

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Part b

This works very similar as Part a. We only give an outline. Note that
$$
(2+\sqrt 2)^2=4+4\sqrt 2+2=6+4\sqrt 2.
$$ Therefore
$$
\sqrt{6+4\sqrt 2}-\sqrt 2=(2+\sqrt 2)-\sqrt 2=2
$$ is rational.