If $f: R \rightarrow R$ is a function defined by $f(x)=[x-1] \cos \left(\frac{2 x-1}{2}\right) \pi,$ where $[.]$ denotes the greatest integer function,then $f$ is

Given $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & \text{if } x < 0 \\ a, & \text{if } x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text{if } x > 0 \end{cases}$
If $f(x)$ is continuous at $x=0$,then the value of $a$ is:

If the function $f(x) = \frac{1-\sin 2x + \cos 2x}{1+\sin 2x + \cos 2x}$ for $x \neq \frac{\pi}{2}$ and $f(x) = k$ for $x = \frac{\pi}{2}$ is continuous at $x = \frac{\pi}{2}$,then $k = $

If $f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ 3, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then the value of $k$ is equal to . . . . . . .

If the function $f(\alpha) = \begin{cases} \frac{1-\cos 6 \alpha}{36 \alpha^2}, & \alpha \neq 0 \\ k, & \alpha=0 \end{cases}$ is continuous at $\alpha=0$,then $k$ is equal to . . . . . . .

If $a$ is the point of discontinuity of the function $f(x) = \begin{cases} \cos 2 x, & \text{for } -\infty < x < 0 \\ e^{3 x}, & \text{for } 0 \leq x < 3 \\ x^2-4 x+3, & \text{for } 3 \leq x \leq 6 \\ \frac{\log (15 x-89)}{x-6}, & \text{for } x>6 \end{cases}$ Then, $\lim _{x \rightarrow a} \frac{x^2-9}{x^3-5 x^2+9 x-9} =$