All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$e^x (1+ \mathrm{sin}2x)$

$= (1 + x + \frac{x^2}{2!} + \ldots)(1+2x-\ldots)$

$= 1 + 3x + \frac{5}{2}x^2 + \ldots$

(ii)
$(1 + \frac{4}{3}x)^n$

$= 1 + n\frac{4}{3}x + \frac{n(n-1)}{2!}(\frac{4}{3}x)^2 + \ldots$

Since the first two terms are equal,

$\frac{4}{3}n = 3 \Rightarrow n = \frac{9}{4}$

Third term $= \frac{n(n-1)}{2!}(\frac{4}{3}x)^2 = \frac{5}{2}x^2$