Set up a coordinate system for the cube.

Let the vertex A be at the origin $(0,0,0)$. Since the cube has side length $a$, the coordinates of the vertices are:
$A = (0,0,0)$
$B = (a,0,0)$
$C = (a,a,0)$
$D = (0,a,0)$
$E = (0,0,a)$
$F = (a,0,a)$
$G = (a,a,a)$
$H = (0,a,a)$

$\vec{AB} = B - A = (a,0,0) - (0,0,0) = (a,0,0)$
$\vec{AE} = E - A = (0,0,a) - (0,0,0) = (0,0,a)$
$\vec{AC} = C - A = (a,a,0) - (0,0,0) = (a,a,0)$
$\vec{CG} = G - C = (a,a,a) - (a,a,0) = (0,0,a)$
$\vec{EG} = G - E = (a,a,a) - (0,0,a) = (a,a,0)$
$\vec{GE} = E - G = (0,0,a) - (a,a,a) = (-a,-a,0)$

The dot product of two vectors $\vec{u} = (u_x, u_y, u_z)$ and $\vec{v} = (v_x, v_y, v_z)$ is given by $\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z$.

1. Calculate $\vec{AB} \cdot \vec{AE}$:

   $$\vec{AB} \cdot \vec{AE} = (a)(0) + (0)(0) + (0)(a) = 0 + 0 + 0 = 0$$

   This is expected as $\vec{AB}$ and $\vec{AE}$ are perpendicular edges of the cube.

2. Calculate $\vec{AB} \cdot \vec{AC}$:

   $$\vec{AB} \cdot \vec{AC} = (a)(a) + (0)(a) + (0)(0) = a^2 + 0 + 0 = a^2$$

3. Calculate $\vec{AB} \cdot \vec{CG}$:

   $$\vec{AB} \cdot \vec{CG} = (a)(0) + (0)(0) + (0)(a) = 0 + 0 + 0 = 0$$

   This is expected as $\vec{AB}$ is parallel to the x-axis and $\vec{CG}$ is parallel to the z-axis, making them perpendicular.

4. Calculate $\vec{AB} \cdot \vec{EG}$:

   $$\vec{AB} \cdot \vec{EG} = (a)(a) + (0)(a) + (0)(0) = a^2 + 0 + 0 = a^2$$

5. Calculate $\vec{AC} \cdot \vec{GE}$:

   $$\vec{AC} \cdot \vec{GE} = (a)(-a) + (a)(-a) + (0)(0) = -a^2 - a^2 + 0 = -2a^2$$

The calculated dot products are:
$\vec{AB} \cdot \vec{AE} = \boxed{\text{0}}$
$\vec{AB} \cdot \vec{AC} = \boxed{a^2}$
$\vec{AB} \cdot \vec{CG} = \boxed{\text{0}}$
$\vec{AB} \cdot \vec{EG} = \boxed{a^2}$
$\vec{AC} \cdot \vec{GE} = \boxed{-2a^2}$