### Part (a)

1. Given Condition:
$$|a - b| = |a + b - 2z|$$
This implies that the distance between $a$ and $b$ is equal to the distance between $a + b$ and $2z$.

2. Equation to Prove:
$$|z - a|^x + |\bar{z} - b|^x = |a - b|^x$$

3. Using the Given Condition:
$$|a - b| = |a + b - 2z|$$
Let $d = |a - b|$. Then:
$$d = |a + b - 2z|$$

4. **Expressing $z$ in Terms of $a$ and $b$:**
Since $z \in \mathbb{C} \setminus \mathbb{R}$, we can write $z = x + yi$ where $y \neq 0$.

5. Using the Triangle Inequality:
$$|z - a| + |z - b| \geq |a - b|$$
Since $z$ is not real, the equality holds only if $z$ lies on the line segment joining $a$ and $b$, which is not possible here.

6. Simplifying the Equation:
$$|z - a|^x + |\bar{z} - b|^x = |a - b|^x$$
Let $f(x) = |z - a|^x + |\bar{z} - b|^x - |a - b|^x$.

7. **Analyzing $f(x)$:**
$$f(x) = 0 \implies |z - a|^x + |\bar{z} - b|^x = |a - b|^x$$
We need to show that $f(x) = 0$ has a unique solution.

8. **Checking $x = 2$:**
$$|z - a|^2 + |\bar{z} - b|^2 = |a - b|^2$$
This is true because:
$$|z - a|^2 + |\bar{z} - b|^2 = |z - a|^2 + |z - b|^2 = |a - b|^2$$
Hence, $x = 2$ is a solution.

9. Uniqueness:
Since $f(x)$ is strictly decreasing, there can be only one solution. Therefore, $x = 2$ is the unique solution.

$\blacksquare$

### Part (b)

1. Given Inequation:
$$|z - a|^x + |\bar{z} - b|^x \leq |a - b|^x$$

2. Analyzing the Function:
From part (a), we know that $f(x) = |z - a|^x + |\bar{z} - b|^x - |a - b|^x$ is strictly decreasing.

3. **Behavior of $f(x)$:**
Since $f(x)$ is strictly decreasing and $f(2) = 0$, for $x \geq 2$, $f(x) \leq 0$.

4. Conclusion:
$$x \geq 2$$

$\blacksquare$

The final answer is $\boxed{ x \geq 2 }$