
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
We know that squaring and square root operations are inverse operations, such as $a^2 \pm 2ab + b^2 = (a \pm b)^2$, then $\sqrt{a^2 \pm 2ab + b^2} = |a \pm b|$. How can a nested radical of the form $\sqrt{a \pm 2\sqrt{b}}$ (where $a > 0$, $b > 0$, and $a \pm 2\sqrt{b} > 0$) be simplified? If two numbers $m, n$ (where $m > 0$, $n > 0$) can be found such that $(\sqrt{m})^2 + (\sqrt{n})^2 = a$ (i.e., $m+n=a$) and $\sqrt{m} \cdot \sqrt{n} = \sqrt{b}$ (i.e., $mn = b$), then $a \pm 2\sqrt{b} = ( (\sqrt{m})^2 + (\sqrt{n})^2 \pm 2\sqrt{m} \cdot \sqrt{n} ) = (\sqrt{m} \pm \sqrt{n})^2$. Therefore, $\sqrt{a \pm 2\sqrt{b}} = |\sqrt{m} \pm \sqrt{n}|$, and the nested radical is simplified. For example, simplify $\sqrt{3+2\sqrt{2}}$; since $3 = 1+2$ and $2 = 1\cdot2$, we have $3+2\sqrt{2} = (\sqrt{1})^2 + (\sqrt{2})^2 + 2\sqrt{1}\cdot\sqrt{2}$, so $\sqrt{3+2\sqrt{2}} = 1 + \sqrt{2}$. By rationalizing any quadratic radical to the form $\sqrt{a \pm 2\sqrt{b}}$ and finding $m, n$ ($m > 0, n > 0$) such that $m+n=a$ and $mn=b$ we can simplify the nested radical. Please read the above material and complete the following questions:
(1) Fill in the blanks: $\sqrt{5 - 2\sqrt{6}} = \_\_\_\_$; $\sqrt{12+2\sqrt{35}} = \_\_\_\_$;
(2) Simplify:
  i. $\sqrt{9 + 6\sqrt{2}}$;
  ii. $\sqrt{16 - 4\sqrt{15}}$;
(3) Compute: $\sqrt{3 - \sqrt{5}} + \sqrt{2 + \sqrt{3}}$.

Present the answer in LaTex format: \boxed{Your answer}
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