The problem asks us to find the integer part $a$ and the fractional part $b$ of the number $\frac{2}{\sqrt{3}-1}$. Then, we need to find the values of the expressions in (1) and (2) involving $a$ and $b$. Specifically, we need to find the value of $b$ in the form $b = \sqrt{\boxed{20-1}} - \boxed{20-2}$ and the value of $b^2 - 2a$ in the form $b^2 - 2a = -\boxed{21-1} \sqrt{\boxed{21-2}}$.

1. Problem Description

The problem asks us to find the integer part $a$ and the fractional part $b$ of the number $\frac{2}{\sqrt{3}-1}$. Then, we need to find the values of the expressions in (1) and (2) involving $a$ and $b$. Specifically, we need to find the value of $b$ in the form $b = \sqrt{\boxed{20-1}} - \boxed{20-2}$ and the value of $b^2 - 2a$ in the form $b^2 - 2a = -\boxed{21-1} \sqrt{\boxed{21-2}}$.

2. Solution Steps

First, let's simplify the expression $\frac{2}{\sqrt{3}-1}$. We can rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{3}+1$:

$$$$

\frac{2}{\sqrt{3}-1} = \frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{2(\sqrt{3}+1)}{3-1} = \frac{2(\sqrt{3}+1)}{2} = \sqrt{3}+1

$$$$

Now, we know that $\sqrt{3}$ is between 1 and 2, since $1^2 = 1 < 3 < 4 = 2^2$. A closer approximation is $\sqrt{3} \approx 1.732$.

Therefore, $\sqrt{3}+1 \approx 1.732 + 1 = 2.732$.

So the integer part $a$ is 2, and the fractional part $b$ is $\sqrt{3}+1 - 2 = \sqrt{3} - 1$.

Now we can find the values for (1) and (2).

(1) We have $b = \sqrt{3}-1$. Comparing this to the given form $b = \sqrt{\boxed{20-1}} - \boxed{20-2}$, we can see that the boxes should be filled with 3 and 1 respectively. So $b = \sqrt{3} - 1$.

(2) Now we need to find $b^2 - 2a$. We have $a = 2$ and $b = \sqrt{3} - 1$.

So $b^2 = (\sqrt{3}-1)^2 = (\sqrt{3})^2 - 2(\sqrt{3})(1) + 1^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}$.

Thus, $b^2 - 2a = (4 - 2\sqrt{3}) - 2(2) = 4 - 2\sqrt{3} - 4 = -2\sqrt{3}$.

Comparing this to the given form $b^2 - 2a = -\boxed{21-1} \sqrt{\boxed{21-2}}$, we see that the boxes should be filled with 2 and 3 respectively. So $b^2 - 2a = -2\sqrt{3}$.

3. Final Answer

(1) $b = \sqrt{3} - 1$

(2) $b^2 - 2a = -2\sqrt{3}$

So, the values to fill in the boxes are:

20-1: 3

20-2: 1

21-1: 2

21-2: 3