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.

We have $$F_{-n}=\frac{(-a_-)^n-(-a_+)^n}{\sqrt5} \tag{1}\label{1}$$ with $$a_\pm:=\frac{1\pm\sqrt5}2.$$

We also have the easy formula for $\sum_{j=j_1}^{j_2}(p+qj)x^j$, which yields
$$\sum_{i=1}^{n-1} \left(\left\lfloor\frac{n-i+1}{2}\right\rfloor + 1\right)u^i \\ =-\frac{(u-1) u^2 \left\lfloor \frac{n+1}{2}\right\rfloor +(u-1) u \left\lfloor \frac{n}{2}\right\rfloor -2 u^{n+2}+u^n+u^3+u^2-u}{(u-1)^2 (u+1)}. \tag{2}\label{2}$$

Your first identity easily follows from \eqref{1} and \eqref{2}.

Other such identities should be derivable quite similarly.