Q.1. Use Walli's formula
http://www.richardhitt.com/courses/227/su99/docs/wallis.pdf
Answer will be $\frac{5.3.1}{8.6.4.2} * \frac{\pi }{2}$
Q.2. $\lim_{x\rightarrow 0} \frac{1 - cos^{2}x}{2x^{4}}$ $\left (\frac{0}{0} form \right )$ $\therefore$ Use L' hospitals rule
So, $\lim_{x\rightarrow 0} \frac{1 - cos^{2}x}{2x^{4}}$
= $\lim_{x\rightarrow 0} \frac{-2cosx*(-sinx)}{8x^{3}}$
= $\lim_{x\rightarrow 0} \frac{sin 2x}{8x^{3}}$
= $\lim_{x\rightarrow 0} \frac{sin 2x}{2x} * \frac{1}{4x^{2}}$
= $\lim_{x\rightarrow 0} 1 * \frac{1}{4x^{2}}$
= infinity
