Assume that we know $A, B, G$ and also we know $y^{(1)}_k = FA^kG$ and $y^{(2)}_k = FA^kB$. But I don't know $F$. I want to find $F$.
I want to take adventages of both $y^{(1)}_k$ and $y^{(2)}_k$, and that's the question. How can I find $F$ if I need to take adventage to $y^{(1)}_k$ and $y^{(2)}_k$? They are measurement data by the way of a impulse response.
You need the Moore-Penrose pseudoinverse inverse:
$$ P^+ \;\; =\;\; P^T\left (PP^T\right )^{-1}. $$
In particular, you should solve for
$$ F \;\; =\;\; y_k^{(1)}\left (A^kG\right )^+ \hspace{2pc}\text{and}\hspace{2pc} F \;\; =\;\; y_k^{(2)}\left (A^kB\right )^+. $$
In terms of "blending" your measurements together to get one answer for $F$, I can see two options:
Averaging
If $y_k^{(1)}$ and $y_k^{(2)}$ were the same size, then you can blend your solution together by adding both equations together:
$$ y_k^{(1)} + y_k^{(2)} \;\; =\;\; FA^k(G+B) $$
in which case you solve
$$ F \;\; =\;\; \left (y_k^{(1)} + y_k^{(2)}\right )\left [A^k(B+G)\right ]^+. $$
Concatenation
If your vectors are not the same size, then what you can do instead is concatenate them to obtain
$$ \left [ \begin{array}{c} y_k^{(1)} \\ y_k^{(2)} \\ \end{array} \right ] \;\; =\;\; \left [ \begin{array}{c} FA^kG \\ FA^kB \\ \end{array} \right ] \;\; \doteq \;\; \widetilde{F}\left [ \begin{array}{c} A^kG \\ A^kB \\ \end{array} \right ] $$
then you can solve for the larger concatenated $\widetilde{F}$ matrix as
$$ \widetilde{F} \;\; =\;\; \left [ \begin{array}{c} y_k^{(1)} \\ y_k^{(2)} \\ \end{array} \right ]\left [ \begin{array}{cc} \left (A^kG\right )^+ & 0 \\ 0 & \left (A^kB\right )^+ \\ \end{array} \right ] $$