$Q_{28}$: $\lim_{x \rightarrow 0}\frac{e^x -(1+ x+\frac{x^2}{2})}{x^3}$
Here, putting $x = 0$ we get $\frac{0}{0}$ form. Appling L'Hospital's Rule (diffrentiating both numerator and denominator independently)
$\lim_{x \rightarrow 0}\frac{e^x -(0+ 1+\frac{2x}{2})}{3x^2}$
$\color{blue}{\lim_{x \rightarrow 0}\frac{e^x}{6} = \frac{1}{6}}$
$Q_{27}$: Let $t = \sqrt{x+\sqrt{x+\sqrt{x +.....\infty}}}$
$t = \sqrt{x+t}$ // t can take only positive values
$t^2 = x+t$
At $x=2$, $t^2 - t - 2 =0$ $\Rightarrow$ $t = 2$
Now $y = x + \sqrt{x+\sqrt{x+\sqrt{x +.....\infty}}}$ i.e,. $y = x + t$
At $x = 2$, $t = 2$ and $\color{blue}{y = 2+ 2 = 4}$
