for a) i get kn/2
for b) BF=3n/4k +n/4
CG=-4n/3k+n
for c) i get k=25/9
1. Let $\displaystyle |\overline{DA}|=k~\implies~|\overline{DG}|=\frac34 k$
$\displaystyle |\overline{CD}|=n~\implies~|\overline{CF}|=\frac34 n$
Then you know: $\displaystyle k \cdot n = 600$
2. Area of triangle $\displaystyle \Delta(CDG)$ is calculated by
$\displaystyle a_{\Delta(CDG)}= \frac12 \cdot |\overline{DG}| \cdot n = \frac12 \cdot \frac34 k \cdot n = \frac38 \cdot k \cdot n = 225$
3. The equations of the lines are OK. Determine the coordinates of the point of intersection, which is J:
$\displaystyle \frac34 \cdot \frac nk + \frac n4 = -\frac43 \cdot \frac nk + n$
Now replace $\displaystyle k = \frac{600}n$ and solve for n.
4. I'll leave the rest for you.