The value of $$4\frac{2}{5}\div\left\{\left(1\frac{1}{2}-3\frac{1}{5}\right)\div3\frac{2}{5}+\left(2\frac{1}{5}\div1\frac{1}{2}+4\frac{2}{5}\right)+\frac{1}{2}\right\}$$ is:

Given Equation : 

$$4\frac{2}{5}\div\left\{\left(1\frac{1}{2}-3\frac{1}{5}\right)\div3\frac{2}{5}+\left(2\frac{1}{5}\div1\frac{1}{2}+4\frac{2}{5}\right)+\frac{1}{2}\right\}$$

Lets solve as per the rule of BODMAS:

= $$\frac{22}{5}\div\left\{\left(\frac{3}{2}-\frac{16}{5}\right)\div\frac{17}{5}+\left(\frac{11}{5}\div\frac{3}{2}+\frac{22}{5}\right)+\frac{1}{2}\right\}$$

= $$\frac{22}{5}\div\left\{\left(-\frac{17}{10}\right)\div\frac{17}{5}+\left(\frac{22}{15}+\frac{22}{5}\right)+\frac{1}{2}\right\}$$ 

= $$\frac{22}{5}\div\left\{\left(-\frac{1}{2}+\left(\frac{22}{15}+\frac{22}{5}\right)+\frac{1}{2}\right)\right\}$$

= $$\frac{22}{5}\div\left\{\frac{88}{15}\right\}$$

= $$\frac{3}{4}$$

Hence, Option D is correct.