The sum of the coefficient of $$x^{2/3}$$ and $$x^{-2/5}$$ in the binomial expansion of $$\left(x^{2/3} + \frac{1}{2}x^{-2/5}\right)^9$$ is

$$\left(x^{2/3}+\frac{1}{2}x^{-2/5}\right)^9$$. General term: $$T_{r+1} = \binom{9}{r}(x^{2/3})^{9-r}\left(\frac{1}{2}x^{-2/5}\right)^r = \binom{9}{r}\frac{1}{2^r}x^{2(9-r)/3-2r/5}$$.

Power of x: $$\frac{2(9-r)}{3} - \frac{2r}{5} = \frac{10(9-r)-6r}{15} = \frac{90-16r}{15}$$.

For $$x^{2/3}$$: $$\frac{90-16r}{15} = \frac{2}{3} = \frac{10}{15}$$. $$90-16r = 10$$. $$r = 5$$.

Coefficient: $$\binom{9}{5}/2^5 = 126/32 = 63/16$$.

For $$x^{-2/5}$$: $$\frac{90-16r}{15} = -\frac{2}{5} = -\frac{6}{15}$$. $$90-16r = -6$$. $$r = 6$$.

Coefficient: $$\binom{9}{6}/2^6 = 84/64 = 21/16$$.

Sum = $$63/16 + 21/16 = 84/16 = 21/4$$.

The correct answer is Option 1: 21/4.