12. Given two quadratic functions $y_{1}$ and $y_{2}$, when $x=\alpha(\alpha>0)$, $y_{1}$ reaches its maximum value of 5, and $y_{2}=25$. Also, the minimum value of $y_{2}$ is $-2$, and $y_{1}+y_{2}=x^{2}+16 x+13$. Find the value of $\alpha$ and the analytical expressions of the quadratic functions $y_{1}, y_{2}$.
Three, 12. Let $y_{1}=m(x-\alpha)^{2}+5$.
Then $y_{2}=x^{2}+16 x+13-m(x-\alpha)^{2}-5$.
When $x=\alpha$, $y_{2}=25$, i.e., $\alpha^{2}+16 \alpha+8=25$.
Solving, we get $\alpha_{1}=1, \alpha_{2}=-17$ (discard).
Thus, $y_{2}=x^{2}+16 x+13-m(x-1)^{2}-5$
$$
=(1-m) x^{2}+(16+2 m) x+(8-m) \text {. }
$$

Also, the minimum value of $y_{2}$ is -2, so
$$
\frac{4(1-m)(8-m)-(16+2 m)^{2}}{4(1-m)}=-2 \text {. }
$$

Solving, we get $m=-2$.
Therefore, $y_{1}=-2 x^{2}+4 x+3, y_{2}=3 x^{2}+12 x+10$.