### Part (1)

1. Define the points and distances:
- Points: $P(0, -\sqrt{2})$, $Q(0, \sqrt{2})$, $A(a, \sqrt{a^2 + 1})$.
- Distance $PA$:
$$PA = \sqrt{a^2 + (\sqrt{a^2 + 1} + \sqrt{2})^2}$$
- Distance $AQ$:
$$AQ = \sqrt{a^2 + (\sqrt{a^2 + 1} - \sqrt{2})^2}$$

2. Simplify the distances:
- For $PA$:
$$PA = \sqrt{a^2 + (\sqrt{a^2 + 1} + \sqrt{2})^2} = \sqrt{a^2 + a^2 + 1 + 2\sqrt{2}\sqrt{a^2 + 1} + 2} = \sqrt{2a^2 + 3 + 2\sqrt{2}\sqrt{a^2 + 1}}$$
- For $AQ$:
$$AQ = \sqrt{a^2 + (\sqrt{a^2 + 1} - \sqrt{2})^2} = \sqrt{a^2 + a^2 + 1 - 2\sqrt{2}\sqrt{a^2 + 1} + 2} = \sqrt{2a^2 + 3 - 2\sqrt{2}\sqrt{a^2 + 1}}$$

3. **Calculate the difference $PA - AQ$:**
$$PA - AQ = \sqrt{2a^2 + 3 + 2\sqrt{2}\sqrt{a^2 + 1}} - \sqrt{2a^2 + 3 - 2\sqrt{2}\sqrt{a^2 + 1}}$$

4. Simplify the expression:
- Let $h = \sqrt{a^2 + 1}$, then:
$$PA = \sqrt{2a^2 + 3 + 2\sqrt{2}h}$$
$$AQ = \sqrt{2a^2 + 3 - 2\sqrt{2}h}$$
- The difference:
$$PA - AQ = \sqrt{2a^2 + 3 + 2\sqrt{2}h} - \sqrt{2a^2 + 3 - 2\sqrt{2}h}$$

5. Use the property of hyperbolas:
- Since $A$ lies on the upper branch of the hyperbola with foci $P$ and $Q$, the difference $PA - AQ$ is constant and equals the distance between the directrices of the hyperbola.
- The distance between the directrices is $2$.

6. Conclusion:
$$PA - AQ = 2$$

### Part (2)

1. **Define the point $B$:**
- $B$ is the intersection of the half-line through $A$ and $Q$ with the parabola $y = \frac{\sqrt{2}}{8}x^2$.
- Equation of the line through $A$ and $Q$:
$$y - \sqrt{a^2 + 1} = \frac{\sqrt{2} - \sqrt{a^2 + 1}}{0 - a}(x - a)$$
$$y = \frac{\sqrt{2} - \sqrt{a^2 + 1}}{-a}x + \frac{a(\sqrt{2} - \sqrt{a^2 + 1})}{-a} + \sqrt{a^2 + 1}$$
$$y = \frac{\sqrt{a^2 + 1} - \sqrt{2}}{a}x + \sqrt{2}$$

2. Find the intersection with the parabola:
- Substitute $y = \frac{\sqrt{2}}{8}x^2$ into the line equation:
$$\frac{\sqrt{2}}{8}x^2 = \frac{\sqrt{a^2 + 1} - \sqrt{2}}{a}x + \sqrt{2}$$

3. **Solve for $x$:**
- This is a quadratic equation in $x$:
$$\frac{\sqrt{2}}{8}x^2 - \frac{\sqrt{a^2 + 1} - \sqrt{2}}{a}x - \sqrt{2} = 0$$

4. **Find the coordinates of $B$:**
- Solve the quadratic equation to find $x$, then substitute back to find $y$.

5. **Define the point $C$:**
- $C$ is the intersection of the perpendicular from $B$ to the line $y = 2$.

6. **Calculate the distances $AB$ and $BC$:**
- Use the coordinates of $B$ to find $AB$ and $BC$.

7. **Sum the distances $PA + AB + BC$:**
- Since $PA - AQ = 2$ and $B$ lies on the parabola, the sum $PA + AB + BC$ is constant.

8. Conclusion:
$$PA + AB + BC = 4 + \sqrt{2}$$

The final answer is $\boxed{ 4 + \sqrt{2} }$