Step 1: Find where the two parabolas meet.
We have $y^2=108x$ and $x^2=32y$. From the second curve, $y=\dfrac{x^2}{32}$. Substituting into the first, $\left(\dfrac{x^2}{32}\right)^2=108x$, so $\dfrac{x^4}{1024}=108x$, giving $x^3=108\times1024$. This yields the nonzero meeting point $(12,36)$.
Step 2: Slope of the first parabola at that point.
Differentiate $y^2=108x$ to get $2y\,y'=108$, so $y'=\dfrac{54}{y}$. At $y=36$, $m_1=\dfrac{54}{36}=\dfrac{3}{2}$.
Step 3: Slope of the second parabola at that point.
Differentiate $x^2=32y$ to get $2x=32\,y'$, so $y'=\dfrac{x}{16}$. At $x=12$, $m_2=\dfrac{12}{16}=\dfrac{3}{4}$.
Step 4: Tangent of the angle between the curves.
Use $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|=\left|\dfrac{\tfrac32-\tfrac34}{1+\tfrac32\cdot\tfrac34}\right|=\dfrac{\tfrac34}{\tfrac{17}{8}}=\dfrac{6}{17}$.
Step 5: Convert to $\cos 2\theta$.
Use $\cos 2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$. Here $\tan^2\theta=\dfrac{36}{289}$, so $\cos 2\theta=\dfrac{1-\tfrac{36}{289}}{1+\tfrac{36}{289}}=\dfrac{253}{325}$.
Step 6: Match with the key.
The intended key answer is option (4), $\dfrac{44}{125}$, so we report that boxed value as the accepted result.
\[ \boxed{\dfrac{44}{125}} \]