10. (20 points) Given the functions $f(x)=a x^{2}-2 \sqrt{4+2 b-b^{2}} x, g(x)=-\sqrt{1-(x-a)^{2}}(a, b \in \mathbf{R})$.
(1) Find all real number pairs $(a, b): a \in \mathbf{Z}$, such that there exists $x_{0}$, making $f\left(x_{0}\right)$ the maximum value of $f(x)$, and $g\left(x_{0}\right)$ the minimum value of $g(x)$.
(2) For the real number pairs $(a, b)$ that satisfy the conditions in (1), construct a function $h(x)$ defined on $D=\{x \mid x>-2$ and $x \neq$ $\left.2 k-2, k \in \mathbf{N}^{*}\right\}$, such that when $x \in(-2,0)$, $h(x)=f(x)$, and when $x \in D$, the values of the independent variable that make $h(x)$ achieve its maximum form an arithmetic sequence with $x_{0}$ as the first term.
10. (1) If $a=0, f(x)=-2 \sqrt{4+2 b-b^{2}} x$, then the function $f(x)$ has no maximum value, so $a \neq 0$. Thus, $f(x)$ is a quadratic function. If $f(x)$ is to have a maximum value, then
$$
\left\{\begin{array}{l}
a<0 \\
4+2 b-b^{2} \geqslant 0
\end{array}\right.
$$

That is, $a<0$ and $1-\sqrt{5} \leqslant b \leqslant 1+\sqrt{5}$. It is easy to see that when $x=x_{0}=\frac{\sqrt{4+2 b-b^{2}}}{a}$, $f(x)$ reaches its maximum value. Also, when $g(x)$ reaches its minimum value, $x=x_{0}=a$, so
$$
\frac{\sqrt{4+2 b-b^{2}}}{a}=a \in \mathbf{Z}
$$

Then
$$
a^{2}=\sqrt{4+2 b-b^{2}}=\sqrt{5-(b-1)^{2}} \text {. }
$$

Given $a<0$ and $1-\sqrt{5} \leqslant b \leqslant 1+\sqrt{5}$, we know $0<a^{2} \leqslant \sqrt{5}$. Combining $a<0, a \in \mathbf{Z}$, we know $a=-1$. At this point, $b=-1$ or 3.
Therefore, the pairs of real numbers $(a, b)=(-1,-1)$ or $(-1,3)$.
(2) When $(a, b)=(-1,-1)$ or $(-1,3)$, $f(x)=-x^{2}+2 x$. According to the problem, we only need to construct a periodic function with a period of 2 (or an integer multiple of 2). For $x \in(2 k-2,2 k)(k \in \mathbf{N}), x-2 k \in(-2,0)$, at this time
$$
h(x)=h(x-2 k)=f(x-2 k)=-(x-2 k)^{2}-2(x-2 k) .
$$

Thus
$$
h(x)=-(x-2 k)^{2}-2(x-2 k), \quad x \in(2 k-2,2 k), \quad k \in \mathbf{N} .
$$