Correct Answer: $\frac{59}{48}$

Solution : Given: $\frac{4 \tan ^2 30^{\circ}+\sin ^2 30^{\circ} \cos ^2 45^{\circ}+\sec ^2 48^{\circ}-\cot ^2 42^{\circ}}{\cos 37^{\circ} \sin 53^{\circ}+\sin 37^{\circ} \cos 53^{\circ}+\tan 18^{\circ} \tan 72^{\circ}}$
We know that $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$
$\sin 30^{\circ} = \frac{1}{2}$
$\cos 45^{\circ} = \frac{1}{\sqrt{2}}$
$\cos 37^{\circ} = \cos (90^{\circ} - 53^{\circ}) = \sin 53^{\circ}$
$\sin 37^{\circ} = \sin (90^{\circ} - 53^{\circ}) = \cos 53^{\circ}$
$\tan 18^{\circ} \tan 72^{\circ} = 1$ (since $18^{\circ}$ and $72^{\circ}$ are complementary angles)
Putting the values, we get:
$= \frac{4 (\frac{1}{\sqrt{3}})^2 + (\frac{1}{2})^2 (\frac{1}{\sqrt{2}})^2 + \sec ^2 48^{\circ} - \tan ^2 48^{\circ}}{\sin 53^{\circ} \sin 53^{\circ} + \cos 53^{\circ} \cos 53^{\circ} + 1}$
$= \frac{4 (\frac{1}{3}) + \frac{1}{8} +1}{(\sin ^2 53^{\circ} + \cos ^2 53^{\circ}) + 1}$
Since $\sin ^2 \theta + \cos ^2 \theta = 1$ for any angle $\theta$, the denominator becomes $1 + 1 = 2$.
So, the expression simplifies to:
$=\frac{4 (\frac{1}{3}) + \frac{1}{8} + 1}{2}$
$=\frac{ (\frac{4}{3}) + \frac{1}{8} + 1}{2}$
$=\frac{ (\frac{32+3+24}{24}) }{2}$
$=(\frac{59}{48})$
Hence, the correct answer is $(\frac{59}{48})$.