Answer
$a = \frac{4}{3}$
$b = -2$
Work Step by Step
$\lim\limits_{x \to 0}(\frac{sin~2x}{x^3}+a+\frac{b}{x^2}) = \lim\limits_{x \to 0}(\frac{sin~2x+ax^3+bx}{x^3}) = \frac{0}{0}$
We can apply L'Hospital's Rule.
$\lim\limits_{x \to 0}(\frac{2~cos~2x+3ax^2+b}{3x^2})$
If this limit is $\frac{0}{0}$, then we can apply L'Hospital's Rule again.
$\lim\limits_{x \to 0}(2~cos~2x+3ax^2+b) = 0$
$2+0+b = 0$
$b = -2$
If $ b = -2$, then the limit is $\frac{0}{0}$, and we can apply L'Hospital's Rule again.
$\lim\limits_{x \to 0}(\frac{-4~sin~2x+6ax}{6x}) = \frac{0}{0}$
We can apply L'Hospital's Rule.
$\lim\limits_{x \to 0}(\frac{-8~cos~2x+6a}{6}) = \frac{-8+6a}{6}$
If $(-8+6a) = 0,$ then this limit is $0$
$-8+6a = 0$
$a = \frac{4}{3}$
If $~~a = \frac{4}{3}~~$ and $~~b = -2,~~$ then $~~\lim\limits_{x \to 0}(\frac{sin~2x}{x^3}+a+\frac{b}{x^2}) = 0$