The value of $f(0)$ so that the function $f(x) = \frac{1 - \cos(1 - \cos x)}{x^4}$ is continuous everywhere is

The value of $k$,for which the function $f(x) = \begin{cases} (\frac{4}{5})^{\frac{\tan 4x}{\tan 5x}}, & 0 < x < \frac{\pi}{2} \\ k + \frac{2}{5}, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,is:

Let $f$ be a differentiable function on the open interval $(a, b)$. Which of the following statements must be true?
$I$. $f$ is continuous on the closed interval $[a, b]$
$II$. $f$ is bounded on the open interval $(a, b)$
$III$. If $a < a_1 < b_1 < b$,and $f(a_1) < 0 < f(b_1)$,then there is a number $c$ such that $a_1 < c < b_1$ and $f(c) = 0$

If $f(x) = \frac{e^{2x} - (1 + 4x)^{1/2}}{\ln(1 - x^2)}$ for $x \neq 0$,then $f$ has

If $f(x) = \begin{cases} \frac{3 \sin(\pi x)}{5x} & x \neq 0 \\ 2K & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $K$ is:

The function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$ is not defined at $x = 0$. The value which should be assigned to $f$ at $x = 0$ so that it is continuous at $x = 0$ is: