The value$(s)$ of $x$ for which the function $f(x) = \begin{cases} 1-x, & x < 1 \\ (1-x)(2-x), & 1 \leq x \leq 2 \\ 3-x, & x > 2 \end{cases}$ fails to be continuous is(are):

Let $f:(0, \pi) \rightarrow \mathbb{R}$ be a function given by
$f(x)=\begin{cases} (\frac{8}{7})^{\frac{\tan 8x}{\tan 7x}}, & 0 < x < \frac{\pi}{2} \\ a-8, & x=\frac{\pi}{2} \\ (1+|\cot x|)^{\frac{b}{a}|\tan x|}, & \frac{\pi}{2} < x < \pi \end{cases}$
Where $a, b \in \mathbb{Z}$. If $f$ is continuous at $x=\frac{\pi}{2}$,then $a^2+b^2$ is equal to ..........

Let the function $f$ be defined by the equation $f(x) = \begin{cases} 3x & \text{if } 0 \le x \le 1 \\ 5 - 3x & \text{if } 1 < x \le 2 \end{cases}$,then:

The number of points where the function $f(x) = \begin{cases} |2x^2 - 3x - 7| & \text{if } x \leq -1 \\ [4x^2 - 1] & \text{if } -1 < x < 1 \\ |x+1| + |x-2| & \text{if } x \geq 1 \end{cases}$ is discontinuous,where $[t]$ denotes the greatest integer $\leq t$,is:

If $f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ nx^3 + m, & x > 1 \end{cases}$
For what integers $m$ and $n$ do $\lim_{x \to 0} f(x)$ and $\lim_{x \to 1} f(x)$ exist?

If $f(x)$ is continuous and $f\left( \frac{9}{2} \right) = \frac{2}{9}$,then $\lim_{x \to 0} f \left( \frac{1 - \cos 3x}{x^2} \right)$ is equal to: