Calculate the value of $\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+....+\frac{F_n}{2^n}+....$

Replace $F_n$ with $F_{n-1}+F_{n-2}$, now you have two series that are both similar to $S$

Hint: $$F_n{={\frac {\Phi ^{n}-\Psi ^{n}}{\sqrt {5}}}={\frac {1}{\sqrt {5}}}\left(\left({\frac {1+{\sqrt {5}}}{2}}\right)^{n}-\left({\frac {1-{\sqrt {5}}}{2}}\right)^{n}\right)}$$

$$S = \sum_{n=1}^{\infty}\dfrac{F_n}{2^n}=\dfrac{1}{\sqrt{5}}\sum_{n=1}^{\infty}\left[\left(\dfrac{\Phi}{2}\right)^n- \left(\dfrac{\Psi}{2}\right)^n\right]$$

Hint: if you know that the generating function for the Fibonacci sequence is:

$\displaystyle \sum_{n=0}^\infty F_nx^n = \frac{x}{1-x-x^2}$

then you can substitute $x=\frac 1 2$ and you immediately have

$\displaystyle \sum_{n=0}^\infty \frac{F_n}{2^n} = \frac{\frac 1 2}{1-\frac 1 2 -\frac 1 4} =2 $

So to answer questions like this quickly, you should learn about generating functions.