To simplify the given expressions, we can go through each one step-by-step:

### Simplifying A:
\[ A = \sqrt{7 + 2\sqrt{6}} \]

We can assume it is of the form \(\sqrt{a} + \sqrt{b}\):
\[ 7 + 2\sqrt{6} = (\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab} \]

From this, we have:
1. \(a + b = 7\)
2. \(2\sqrt{ab} = 2\sqrt{6} \Rightarrow \sqrt{ab} = \sqrt{6} \Rightarrow ab = 6\)

Now we can solve the equations:
- We set \(x = a\) and \(y = b\), so \(x + y = 7\) and \(xy = 6\).
- The values can be found by solving the quadratic: \(t^2 - (x+y)t + xy = 0 \Rightarrow t^2 - 7t + 6 = 0\).
- The roots are:
\[ t = \frac{7 \pm 1}{2} = 6 \text{ and } 1 \]
Thus, \(a = 6\) and \(b = 1\).

Finally, we have:
\[ A = \sqrt{6} + \sqrt{1} = \sqrt{6} + 1 \]

### Simplifying C:
\[ C = \sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}} \]

For \( \sqrt{5 + 2\sqrt{6}} \):
- Let’s assume it is of the form \(\sqrt{a} + \sqrt{b}\):
1. \(a + b = 5\)
2. \(2\sqrt{ab} = 2\sqrt{6} \Rightarrow ab = 6\)

Using the quadratic equation:
\[ t^2 - 5t + 6 = 0 \]
Roots are:
\[ t = \frac{5 \pm 1}{2} = 3 \text{ and } 2 \]
So, \(a = 3\) and \(b = 2\),
Thus:
\[ \sqrt{5 + 2\sqrt{6}} = \sqrt{3} + \sqrt{2} \]

For \( \sqrt{5 - 2\sqrt{6}} \):
- It can be similarly shown to give:
\[ \sqrt{5 - 2\sqrt{6}} = \sqrt{3} - \sqrt{2} \]

Combining:
\[ C = (\sqrt{3} + \sqrt{2}) + (\sqrt{3} - \sqrt{2}) = 2\sqrt{3} \]

### Simplifying B:
\[ B = \sqrt{63 - 36\sqrt{3}} \]

Assuming it is of the form \(\sqrt{a} - \sqrt{b}\):
1. \(a + b = 63\)
2. \(2\sqrt{ab} = 36\sqrt{3} \Rightarrow \sqrt{ab} = 18\sqrt{3} \Rightarrow ab = 972\)

Using the quadratic:
\[ t^2 - 63t + 972 = 0 \]
The discriminant is:
\[ \Delta = 63^2 - 4 \cdot 972 = 3969 - 3888 = 81 \]
Thus, roots:
\[ t = \frac{63 \pm 9}{2} = 36 \text{ and } 27 \]

Finally, we have:
\[ B = \sqrt{36} - \sqrt{27} = 6 - 3\sqrt{3} \]

### Summary:
1. \( A = \sqrt{6} + 1 \)
2. \( C = 2\sqrt{3} \)
3. \( B = 6 - 3\sqrt{3} \)