Let's simplify the expressions one by one:

### a)
\[
\sqrt{6} - 2\sqrt{5} - \sqrt{(\sqrt{5} + 1)^2}
\]
First, let's simplify the expression inside the square root:
\[
\sqrt{(\sqrt{5} + 1)^2} = \sqrt{5} + 1
\]
So the expression becomes:
\[
\sqrt{6} - 2\sqrt{5} - (\sqrt{5} + 1)
\]
Now, combine the terms:
\[
\sqrt{6} - 2\sqrt{5} - \sqrt{5} - 1 = \sqrt{6} - 3\sqrt{5} - 1
\]

### b)
\[
\frac{1}{5}\sqrt{75} + \frac{3}{\sqrt{3}} - \sqrt{48}
\]
First, simplify each term:
\[
\sqrt{75} = 5\sqrt{3} \implies \frac{1}{5}\sqrt{75} = \frac{1}{5}(5\sqrt{3}) = \sqrt{3}
\]
For the second term:
\[
\frac{3}{\sqrt{3}} = 3 \cdot \frac{\sqrt{3}}{3} = \sqrt{3}
\]
For the third term:
\[
\sqrt{48} = 4\sqrt{3}
\]
So, the expression becomes:
\[
\sqrt{3} + \sqrt{3} - 4\sqrt{3} = 2\sqrt{3} - 4\sqrt{3} = -2\sqrt{3}
\]

### c)
\[
\frac{2}{\sqrt{3} - 1} - \sqrt{3} + 1
\]
To simplify \(\frac{2}{\sqrt{3} - 1}\), multiply numerator and denominator by \(\sqrt{3} + 1\):
\[
\frac{2(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{2(\sqrt{3} + 1)}{3 - 1} = \frac{2(\sqrt{3} + 1)}{2} = \sqrt{3} + 1
\]
Now substitute back:
\[
\sqrt{3} + 1 - \sqrt{3} + 1 = 2
\]

### d)
\[
3\sqrt{27} + \frac{1}{3}\sqrt{12} - 2\sqrt{3}
\]
Simplify:
\[
\sqrt{27} = 3\sqrt{3} \implies 3\sqrt{27} = 3(3\sqrt{3}) = 9\sqrt{3}
\]
\[
\sqrt{12} = 2\sqrt{3} \implies \frac{1}{3}\sqrt{12} = \frac{1}{3}(2\sqrt{3}) = \frac{2}{3}\sqrt{3}
\]
Combining terms:
\[
9\sqrt{3} + \frac{2}{3}\sqrt{3} - 2\sqrt{3} = (9 - 2)\sqrt{3} + \frac{2}{3}\sqrt{3} = 7\sqrt{3} + \frac{2}{3}\sqrt{3}
\]
Convert \(7\sqrt{3}\) to a fraction:
\[
= \frac{21}{3}\sqrt{3} + \frac{2}{3}\sqrt{3} = \frac{23}{3}\sqrt{3}
\]

### e)
\[
\sqrt{4 + 2\sqrt{3}} - \sqrt{4 - 2\sqrt{3}}
\]
Notice \(4 + 2\sqrt{3} = (\sqrt{3} + 1)^2\) and \(4 - 2\sqrt{3} = (\sqrt{3} - 1)^2\):
\[
\sqrt{4 + 2\sqrt{3}} = \sqrt{(\sqrt{3}+1)^2} = \sqrt{3} + 1
\]
\[
\sqrt{4 - 2\sqrt{3}} = \sqrt{(\sqrt{3}-1)^2} = \sqrt{3} - 1
\]
Substituting back:
\[
(\sqrt{3} + 1) - (\sqrt{3} - 1) = 1 + 1 = 2
\]

### f)
\[
(4 + \sqrt{15})(\sqrt{10} - \sqrt{6})(\sqrt{4} - \sqrt{15})
\]
Calculating each multiplication step:
1. Calculate \((4 + \sqrt{15})(\sqrt{10} - \sqrt{6})\):
- \(4\sqrt{10} - 4\sqrt{6} + \sqrt{15}\sqrt{10} - \sqrt{15}\sqrt{6}\)

2. Simplify \(\sqrt{15}\sqrt{10} = \sqrt{150} = 5\sqrt{6}\) and \(\sqrt{15}\sqrt{6} = \sqrt{90} = 3\sqrt{10}\):
- Resulting in:
\[
4\sqrt{10} - 4\sqrt{6} + 5\sqrt{6} - 3\sqrt{10} = (4 - 3)\sqrt{10} + (5 - 4)\sqrt{6} = \sqrt{10} + \sqrt{6}
\]

3. Now multiply by \((\sqrt{4} - \sqrt{15}) = (2 - \sqrt{15})\):
\[
(\sqrt{10} + \sqrt{6})(2 - \sqrt{15}) = 2\sqrt{10} - \sqrt{10}\sqrt{15} + 2\sqrt{6} - \sqrt{6}\sqrt{15}
\]
Thus:
\[
-\sqrt{150} = -5\sqrt{6}, -\sqrt{90} = -3\sqrt{10}
\]
Final expression:
\[
2\sqrt{10} - 5\sqrt{6} + 2\sqrt{6} - 3\sqrt{10} = (2 - 3)\sqrt{10} + (-5 + 2)\sqrt{6} = -\sqrt{10} - 3\sqrt{6}
\]

### Final Answers:
- a) \(\sqrt{6} - 3\sqrt{5} - 1\)
- b) \(-2\sqrt{3}\)
- c) \(2\)
- d) \(\frac{23}{3}\sqrt{3}\)
- e) \(2\)
- f) \(-\sqrt{10} - 3\sqrt{6}\)