1. **Simplify** $\frac{1}{2} \left( \frac{1}{\sqrt{3}} + \sqrt{2} - \frac{1}{\sqrt{3}} - \sqrt{2} \right)$. Step 1: Notice that $\frac{1}{\sqrt{3}}$ and $-\frac{1}{\sqrt{3}}$ cancel out, and $\sqrt{2}$ and $-\sqrt{2}$ also cancel out. Step 2: So the expression inside the parentheses is $0$. Step 3: Multiply by $\frac{1}{2}$: $\frac{1}{2} \times 0 = 0$. **Answer:** $0$. 2. **Find** $a$ and $b$ if $a\sqrt{6} + b\sqrt{5}\sqrt{2} = 3a\sqrt{10}$. Step 1: Simplify $b\sqrt{5}\sqrt{2} = b\sqrt{10}$. Step 2: Rewrite the equation as $a\sqrt{6} + b\sqrt{10} = 3a\sqrt{10}$. Step 3: Rearrange terms: $a\sqrt{6} = 3a\sqrt{10} - b\sqrt{10} = (3a - b)\sqrt{10}$. Step 4: For the equation to hold, the terms with different surds must be zero separately because $\sqrt{6}$ and $\sqrt{10}$ are irrational and linearly independent. So, coefficient of $\sqrt{6}$: $a = 0$. Step 5: Substitute $a=0$ into the equation: $0 + b\sqrt{10} = 0$ implies $b=0$. **Answer:** $a=0$, $b=0$. 3. **Simplify** $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$. Step 1: Multiply numerator and denominator by the conjugate of the denominator $\sqrt{5} + \sqrt{3}$: $$\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{(\sqrt{5} + \sqrt{3})^2}{(\sqrt{5})^2 - (\sqrt{3})^2}$$ Step 2: Expand numerator: $$(\sqrt{5})^2 + 2\sqrt{5}\sqrt{3} + (\sqrt{3})^2 = 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15}$$ Step 3: Simplify denominator: $$5 - 3 = 2$$ Step 4: So the expression is: $$\frac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15}$$ **Answer:** $4 + \sqrt{15}$.