Let $f(x) = x^3$,$x \in [-1, 1]$. Then which of the following are correct?

Let $f(x)=\lim _{\theta \rightarrow 0}\left(\frac{\cos \pi x-x^{\left(\frac{2}{\theta}\right)} \sin (x-1)}{1+x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x \in R$. Consider the following two statements: $(I)$ $f(x)$ is discontinuous at $x=1$. $(II)$ $f(x)$ is continuous at $x=-1$. Then,

Let $f :[0,1] \rightarrow \mathbb{R}$ and $g :[0,1] \rightarrow \mathbb{R}$ be defined as follows:
$f(x) = \begin{cases} 1 & \text{if } x \text{ is rational} \\ 0 & \text{if } x \text{ is irrational} \end{cases}$
$g(x) = \begin{cases} 0 & \text{if } x \text{ is rational} \\ 1 & \text{if } x \text{ is irrational} \end{cases}$
Then:

Find all points of discontinuity of $f,$ where $f$ is defined by
$f(x) = \begin{cases} |x| + 3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \ge 3 \end{cases}$

Define $f(x) = \begin{cases} \frac{1-\sin x}{(\pi-2x)^2} & \text{, if } x \neq \frac{\pi}{2} \\ k & \text{, if } x = \frac{\pi}{2} \end{cases}$. If $f(x)$ is continuous at $x = \frac{\pi}{2}$,then $k =$

If the function $f(x)$ is continuous in $0 \leq x \leq \pi$,then the value of $2a+3b$ is where $f(x) = \begin{cases} x+a \sqrt{2} \sin x & \text{if } 0 \leq x < \frac{\pi}{4} \\ 2x \cot x + b & \text{if } \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x & \text{if } \frac{\pi}{2} < x \leq \pi \end{cases}$