We have
$$F_{-n}=\frac{(-a_-)^n-(-a_+)^n}{\sqrt5}$$
with
$$a_\pm:=\frac{1\pm\sqrt5}2$$
and the easy formula for $\sum_{j=j_1}^{j_2}(p+qj)x^j$. 

It follows that  
$$\sum_{i=1}^{n-1} \left(\left\lfloor\frac{n-i+1}{2}\right\rfloor + 1\right)u^i \\ 
=-\frac{(u-1) u^2 \left\lfloor \frac{n+1}{2}\right\rfloor +(u-1) u \left\lfloor
   \frac{n}{2}\right\rfloor -2 u^{n+2}+u^n+u^3+u^2-u}{(u-1)^2 (u+1)},$$
from which your first identity easily follows. 

Other such identities should be derived quite similarly.