The characteristic equation of our linear recurrence is $x^{2}-\frac{5}{2} x+1=0$, with roots $x_{1}=2$ and $x_{2}=\frac{1}{2}$. So $F(n)=a \cdot 2^{n}+b \cdot\left(\frac{1}{2}\right)^{n}$ with some constants $a, b$. By $F(0)=0$ and $F(1)=\frac{3}{2}$, these constants satisfy $a+b=0$ and $2 a+\frac{b}{2}=\frac{3}{2}$. So $a=1$ and $b=-1$, and therefore $$F(n)=2^{n}-2^{-n}$$ Observe that $$\frac{1}{F\left(2^{n}\right)}=\frac{2^{2^{n}}}{\left(2^{2^{n}}\right)^{2}-1}=\frac{1}{2^{2^{n}}-1}-\frac{1}{\left(2^{2^{n}}\right)^{2}-1}=\frac{1}{2^{2^{n}}-1}-\frac{1}{2^{2^{n+1}}-1}$$ so $$\sum_{n=0}^{\infty} \frac{1}{F\left(2^{n}\right)}=\sum_{n=0}^{\infty}\left(\frac{1}{2^{2^{n}}-1}-\frac{1}{2^{2^{n+1}}-1}\right)=\frac{1}{2^{2^{0}}-1}=1$$ Hence the sum takes the value 1, which is rational.