Step-by-step Solution:

1. Analyze the nature of the charges:

   • Electric field lines originate from positive charges and terminate on negative charges.
   • In the given diagram, the electric field lines are shown originating from charge $Q_1$ and terminating on charge $Q_2$.
   • This indicates that $Q_1$ is a positive charge ($Q_1 > 0$) and $Q_2$ is a negative charge ($Q_2 < 0$).

2. Compare the magnitudes of the charges ($|Q_1|$ and $|Q_2|$):

   • The density of electric field lines (or the number of lines drawn) originating from or terminating on a charge is proportional to the magnitude of that charge.
   • By counting the number of field lines in the diagram, we can compare the magnitudes of the charges.
   • Number of lines originating from $Q_1$ is 8.
   • Number of lines terminating on $Q_2$ is 4.
   • Since the number of lines associated with $Q_1$ is greater than the number of lines associated with $Q_2$, we can conclude that the magnitude of $Q_1$ is greater than the magnitude of $Q_2$.
   • Therefore, $|Q_1| > |Q_2|$.
   • This makes Option A correct and Option B incorrect.

3. Locate the neutral point (where the electric field is zero):

   • A neutral point occurs where the net electric field due to the system of charges is zero. This happens when the electric fields from the individual charges are equal in magnitude and opposite in direction.
   • Let the position of $Q_1$ be at $x=0$ and $Q_2$ be at $x=d$. The electric field due to $Q_1$ is $\vec{E_1}$ and due to $Q_2$ is $\vec{E_2}$. The net field is $\vec{E} = \vec{E_1} + \vec{E_2}$.
   • Case 1: Between the charges ($0 < x < d$):
     • $\vec{E_1}$ points to the right (away from positive $Q_1$).
     • $\vec{E_2}$ also points to the right (towards negative $Q_2$).
     • Since both fields are in the same direction, they cannot cancel out. So, no neutral point exists between the charges.
   • Case 2: To the left of $Q_1$ ($x < 0$):
     • Let's consider a point at a distance $r_1$ from $Q_1$ and $r_2$ from $Q_2$. So, $r_2 = r_1 + d$.
     • $\vec{E_1}$ points to the left, and its magnitude is $E_1 = \frac{k|Q_1|}{r_1^2}$.
     • $\vec{E_2}$ points to the right, and its magnitude is $E_2 = \frac{k|Q_2|}{r_2^2} = \frac{k|Q_2|}{(r_1+d)^2}$.
     • For the net field to be zero, we need $E_1 = E_2$. However, we know $|Q_1| > |Q_2|$ and for any point to the left of $Q_1$, the distance $r_1 < r_2$. Since the point is closer to the larger charge, the electric field from the larger charge ($E_1$) will always be greater than the electric field from the smaller charge ($E_2$). Thus, they can never cancel.
     • Mathematically: $\frac{E_1}{E_2} = \frac{|Q_1|/r_1^2}{|Q_2|/r_2^2} = \frac{|Q_1|}{|Q_2|} \left(\frac{r_2}{r_1}\right)^2$. Since $\frac{|Q_1|}{|Q_2|} > 1$ and $\frac{r_2}{r_1} > 1$, the ratio $\frac{E_1}{E_2}$ is always greater than 1. So, $E_1 \neq E_2$.
     • Therefore, the electric field cannot be zero at any finite distance to the left of $Q_1$. Option C is incorrect.
   • Case 3: To the right of $Q_2$ ($x > d$):
     • Let's consider a point at a distance $r_2$ from $Q_2$ and $r_1$ from $Q_1$. So, $r_1 = r_2 + d$.
     • $\vec{E_1}$ points to the right, and its magnitude is $E_1 = \frac{k|Q_1|}{r_1^2} = \frac{k|Q_1|}{(r_2+d)^2}$.
     • $\vec{E_2}$ points to the left, and its magnitude is $E_2 = \frac{k|Q_2|}{r_2^2}$.
     • The fields are in opposite directions. For them to cancel, their magnitudes must be equal: $E_1 = E_2$.
     • $\frac{k|Q_1|}{(r_2+d)^2} = \frac{k|Q_2|}{r_2^2}$
     • $\frac{|Q_1|}{|Q_2|} = \frac{(r_2+d)^2}{r_2^2} = \left(1 + \frac{d}{r_2}\right)^2$
     • Since $|Q_1| > |Q_2|$, the ratio $\frac{|Q_1|}{|Q_2|}$ is greater than 1. The equation $\left(1 + \frac{d}{r_2}\right)^2 = \frac{|Q_1|}{|Q_2|}$ has a real, finite, positive solution for $r_2$.
     • This means there exists a neutral point at a finite distance to the right of $Q_2$. Option D is correct.

Conclusion:

Based on the analysis, the correct statements are $|Q_1| > |Q_2|$ and that the electric field is zero at a finite distance to the right of $Q_2$.