For the function $f(x) = \frac{1}{x + 2^{\frac{1}{x - 2}}}$,$x \neq 2$,which of the following holds?

Let $f(x) = \begin{cases} \frac{(x - 1)(6x - 1)}{2x - 1}, & \text{if } x \neq \frac{1}{2} \\ 0, & \text{if } x = \frac{1}{2} \end{cases}$. Then at $x = \frac{1}{2}$,

Let $f: R \rightarrow R$ be a continuous function such that $f(x^2) = f(x^3)$ for all $x \in R$. Consider the following statements:
$I.$ $f$ is an odd function.
$II.$ $f$ is an even function.
$III.$ $f$ is differentiable everywhere.
Then,

Let a function $f: R \rightarrow R$ be defined as
$f(x) = \begin{cases} \sin x - e^x & \text{if } x \leq 0 \\ a + [-x] & \text{if } 0 < x < 1 \\ 2x - b & \text{if } x \geq 1 \end{cases}$
where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous on $R$,then $(a + b)$ is equal to:

The value of $f(0)$ so that the function $f(x) = \frac{2^x - 2^{-x}}{x}$ for $x \neq 0$ is continuous at $x = 0$ is:

Let $f(x) = [x]\sin \left( \frac{\pi}{[x + 1]} \right)$,where $[.]$ denotes the greatest integer function. The domain of $f$ is and the points of discontinuity of $f$ in the domain are