Let's solve each of these equations step by step.

### 1. \( x^2 - 2 = 0 \)

To solve for \( x \):

\[ x^2 - 2 = 0 \]
\[ x^2 = 2 \]
\[ x = \pm\sqrt{2} \]

So, the solutions are \( x = \sqrt{2} \) and \( x = -\sqrt{2} \).

### 2. \( 4x^2 - 3 = 0 \)

To solve for \( x \):

\[ 4x^2 - 3 = 0 \]
\[ 4x^2 = 3 \]
\[ x^2 = \frac{3}{4} \]
\[ x = \pm\sqrt{\frac{3}{4}} \]
\[ x = \pm\frac{\sqrt{3}}{2} \]

So, the solutions are \( x = \frac{\sqrt{3}}{2} \) and \( x = -\frac{\sqrt{3}}{2} \).

### 3. \( x^2 + \frac{2}{3}x + 3 = 0 \)

This is a quadratic equation of the form \( ax^2 + bx + c = 0 \). We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = \frac{2}{3} \), and \( c = 3 \).

\[ x = \frac{-\frac{2}{3} \pm \sqrt{\left(\frac{2}{3}\right)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \]
\[ x = \frac{-\frac{2}{3} \pm \sqrt{\frac{4}{9} - 12}}{2} \]
\[ x = \frac{-\frac{2}{3} \pm \sqrt{\frac{4}{9} - \frac{108}{9}}}{2} \]
\[ x = \frac{-\frac{2}{3} \pm \sqrt{\frac{4 - 108}{9}}}{2} \]
\[ x = \frac{-\frac{2}{3} \pm \sqrt{\frac{-104}{9}}}{2} \]
\[ x = \frac{-\frac{2}{3} \pm \frac{\sqrt{-104}}{3}}{2} \]
\[ x = \frac{-\frac{2}{3} \pm \frac{2i\sqrt{26}}{3}}{2} \]
\[ x = \frac{-2 \pm 2i\sqrt{26}}{6} \]
\[ x = \frac{-1 \pm i\sqrt{26}}{3} \]

So, the solutions are \( x = \frac{-1 + i\sqrt{26}}{3} \) and \( x = \frac{-1 - i\sqrt{26}}{3} \).

### 4. \( x^2 - \frac{2}{2}x + 2 = 0 \)

Simplify the equation first:

\[ x^2 - \frac{2}{2}x + 2 = 0 \]
\[ x^2 - x + 2 = 0 \]

Again, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = 2 \).

\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \]
\[ x = \frac{1 \pm \sqrt{1 - 8}}{2} \]
\[ x = \frac{1 \pm \sqrt{-7}}{2} \]
\[ x = \frac{1 \pm i\sqrt{7}}{2} \]

So, the solutions are \( x = \frac{1 + i\sqrt{7}}{2} \) and \( x = \frac{1 - i\sqrt{7}}{2} \).