Explanation

(a)
Given, $\lim_{x \rightarrow \infty}{f(x) = 1}$
$\Rightarrow \ \ \ \lim_{x \rightarrow \infty} = \frac{ax + b}{x + 1} = 1$
$\Rightarrow \ \ \ \ \ \ \lim_{x \rightarrow \infty}\frac{a + \frac{b}{x}}{1 + \frac{1}{x}} = 1$
$\Rightarrow \ \ \ \ \ \ \ a = 1$
Also, $\lim_{x \rightarrow 0}f(x) = 2$
$\Rightarrow \ \ \ \ \ \ \ \ \ \lim_{x \rightarrow 0}\frac{ax + b}{x + 1} = 2$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ b = 2$
Now, $f( - 2) = \frac{a( - 2) + b}{( - 2) + 1}$
$= \frac{- 2 + 2}{- 2 + 1} = 0$