The first one can be solved using the fact that the generating function of the Fibonacci numbers is $$\frac{z}{1-z-z^2}.$$ Introduce the function $$f(z) = \sum_{n\ge 0} z^n \sum_{k=0}^n \frac{F_{2k} F_{n-k}}{10^n}$$ so that we are interested in $f(1).$
Re-write $f(z)$ as follows: $$f(z) = \sum_{k\ge 0} F_{2k} \sum_{n\ge k} \frac{z^n}{10^n} F_{n-k} = \sum_{k\ge 0} F_{2k} \frac{z^k}{10^k} \sum_{n\ge 0} \frac{z^n}{10^n} F_n.$$
Now we have $$ \sum_{k\ge 0} F_{2k} z^{2k} = \frac{1}{2} \frac{z}{1-z-z^2} - \frac{1}{2} \frac{z}{1+z-z^2}$$ and therefore $f(1)$ is $$\left(\frac{1}{2} \frac{1/\sqrt{10}}{1-1/\sqrt{10}-1/10} - \frac{1}{2} \frac{1/\sqrt{10}}{1+1/\sqrt{10}-1/10}\right) \times \frac{1/10}{1-1/10-1/100}$$ which simplifies to $$\frac{1}{2\sqrt{10}} \frac{2/\sqrt{10}}{81/100-1/10} \times \frac{10}{89} = \frac{1}{10} \times \frac{1}{71/100} \times \frac{10}{89} = \frac{100}{89\times 71}.$$