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

(i)
$f(x) = \mathrm{ln} (1+2sinx)$

$f'(x) = \frac{2cosx}{1+2sinx}$

$f''(x) = \frac{-2sinx(1+2sinx)- 2cosx(2cosx)}{(1+2sinx)^{2}}$

$= \frac{-2sinx-4sin^{2}x- 4cos^{2}x}{(1+2sinx)^{2}}$

$= \frac{-2sinx-4}{(1+2sinx)^{2}}$

$f \prime \prime \prime (x) = \frac{[-2cosx(1+2sinx)^{2} - (-2sinx -4)(1+2sinx)(2)(2cosx)]}{(1+2sinx)^{4}}$

$f(0) = 0$

$f'(0) = 2$

$f''(0) = -4$

$f \prime \prime \prime (0) = 14$

$\therefore,~ f(x) = 2x - 2x^{2} +\frac{7}{3} x^{3} + \ldots$

(ii)
Using the series in MF15,
$e^{ax} sin nx = (1+ ax + \frac{a^{2}x^{2}}{2!}+ \ldots)(nx - \frac{n^{3}x^{3}}{3!}+ \ldots)$

$e^{ax} sin nx = nx - \frac{n^{3}x^{3}}{6}+ anx^{2} + \frac{a^{2}nx^{3}}{2} + \ldots)$

$\therefore, n = 2, an = -2 \Rightarrow a = -1$

Third non-zero term $= [- \frac{2^{3}}{6}+ \frac{(-1)^{2}(2)}{2}]x^{3} = -\frac{8}{6} + 1 = -\frac{1}{3}x^{3}$