As a complement to Iosif's answer, I'll give the inductive proof. First, some preliminaries: note that by recursively applying the defining relations,
$$
F_n=F_{n-1}+F_{n-3}+...+F_2+1=1-(F_{-(n-1)}+F_{-(n-3)}+...+F_{-2}),\text{ when }n\text{ is odd,}
$$
and
$$
F_n=F_{n-1}+F_{n-3}+...+F_1=F_{-(n-1)}+F_{-(n-3)}+...+F_{-1},\text{ when }n\text{ is even.}
$$
Therefore,
$$
F_{-1}+F_{-2}+...+F_{-(n-1)}=1+(-1)^{n-1}F_{n-1}+(-1)^{n}F_{n}=1-F_{-(n-1)}-F_{-n}. \tag{1}\label{1}
$$

Now, we start the inductive proof of your first statement. Clearly it holds for $n=1,2$.

Suppose that $n>2$ is odd. By equation \eqref{1} and the inductive hypothesis, we obtain
\begin{eqnarray*}
F_{-n} &=& F_{-(n-2)} + \left(1-2F_{-(n-1)}-2F_{-(n-2)}-\sum_{i=1}^{n-3}F_{-i}\right)\\
&=&\left(\left\lfloor \frac{n-1}{2} \right\rfloor - \sum_{i=1}^{n-3}\left( \left\lfloor \frac{n-i-1}{2}\right\rfloor +1 \right)F_{-i}\right) + ... \\ 
&&\text{        ...} +\left(1-2F_{-(n-1)}-2F_{-(n-2)}-\sum_{i=1}^{n-3}F_{-i}\right) \\
&=&\left\lfloor \frac{n+1}{2} \right\rfloor - \sum_{i=1}^{n-1}\left( \left\lfloor \frac{n-i+1}{2}\right\rfloor +1 \right)F_{-i}
\end{eqnarray*}
as required. The cases of $n$ even, and the second of your formulae, are similar.