### Part (i)
1. Define the coordinates:
- Let $B$ be at the origin $(0, 0)$.
- Let $A$ be at $(0, 1)$.
- Let $C$ be at $(1, 0)$.
- Let $D$ be at $(1, 1)$.

2. Find the midpoints:
- $M$ is the midpoint of $AB$, so $M = \left(0, \frac{1}{2}\right)$.
- $N$ is the midpoint of $BC$, so $N = \left(\frac{1}{2}, 0\right)$.

3. **Find the equations of the lines $CM$ and $DN$:**
- The line $CM$ passes through $C(1, 0)$ and $M\left(0, \frac{1}{2}\right)$.
- Slope of $CM$ is $\frac{\frac{1}{2} - 0}{0 - 1} = -\frac{1}{2}$.
- Equation of $CM$ is $y = -\frac{1}{2}x + \frac{1}{2}$.
- The line $DN$ passes through $D(1, 1)$ and $N\left(\frac{1}{2}, 0\right)$.
- Slope of $DN$ is $\frac{0 - 1}{\frac{1}{2} - 1} = 2$.
- Equation of $DN$ is $y = 2x - 1$.

4. **Find the intersection point $I$ of $CM$ and $DN$:**
- Set the equations equal: $-\frac{1}{2}x + \frac{1}{2} = 2x - 1$.
- Solve for $x$:
$$-\frac{1}{2}x + \frac{1}{2} = 2x - 1 \implies \frac{1}{2} + 1 = 2x + \frac{1}{2}x \implies \frac{3}{2} = \frac{5}{2}x \implies x = \frac{3}{5}$$
- Substitute $x = \frac{3}{5}$ into $y = 2x - 1$:
$$y = 2 \left(\frac{3}{5}\right) - 1 = \frac{6}{5} - 1 = \frac{1}{5}$$
- So, $I = \left(\frac{3}{5}, \frac{1}{5}\right)$.

5. **Calculate the area of triangle $CIN$:**
- Vertices of $\triangle CIN$ are $C(1, 0)$, $I\left(\frac{3}{5}, \frac{1}{5}\right)$, and $N\left(\frac{1}{2}, 0\right)$.
- Use the formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$:
$$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$
- Substitute the coordinates:
$$\text{Area} = \frac{1}{2} \left| 1\left(\frac{1}{5} - 0\right) + \frac{3}{5}\left(0 - 0\right) + \frac{1}{2}\left(0 - \frac{1}{5}\right) \right| = \frac{1}{2} \left| \frac{1}{5} - \frac{1}{10} \right| = \frac{1}{2} \left| \frac{1}{10} \right| = \frac{1}{20}$$

### Part (ii)
1. **Consider the parallelogram created by connecting $AN, BP, CQ,$ and $DM$:**
- The midpoints $M, N, P, Q$ divide the sides of the parallelogram into equal segments.
- Each midpoint is joined to the two vertices not on its side, forming an 8-pointed star.

2. Calculate the area of the parallelogram formed by the midpoints:
- The area of the parallelogram formed by the midpoints is $\frac{1}{5}$ of the area of the original parallelogram.

3. Calculate the area outside the 8-pointed star:
- The area outside the 8-pointed star is $\frac{4}{5}$ of the area of the original parallelogram.
- The area outside the resulting 8-pointed star is $\frac{2}{5}$ of the area of the parallelogram.

### Part (iii)
1. Use the property of the centroid:
- The centroid $G$ of a triangle divides each median into a ratio of $2:1$.

2. **Calculate the area of $\triangle AGB$:**
- Since $G$ divides the medians in a $2:1$ ratio, the area of $\triangle AGB$ is $\frac{1}{3}$ of the area of $\triangle ABC$.

### Part (iv)
1. Consider the general case for convex quadrilaterals:
- The same logic used in (ii) can be applied to any convex quadrilateral.
- The area outside the resulting 8-pointed star is $\frac{2}{5}$ of the area of the quadrilateral.

The final answer is $\boxed{ \frac{1}{20} }$ for part (i), $\frac{2}{5}$ for part (ii), $\frac{1}{3}$ for part (iii), and True for part (iv).