Example 4 (2005 Guangdong College Entrance Examination Question) Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies: for any real number $x$, $f(2+x)=f(2-x)$, $f(7+x)=f(7-x)$, and in the interval $[0,7]$, only $f(1)=f(3)=0$.
(1) Determine the odd or even nature of $f(x)$;
(2) Find the number of roots of the equation $f(x)=0$ in the interval $[-2005,2005]$.

(1) For $f(2+x)=f(2-x)$, substituting $x+2$ for $x$ gives $f(4+x)=f(-x)$. Similarly, from $f(7+x)=f(7-x)$, we get $f(14+x)=f(-x)$. Therefore, $f(4+x)=f(14+x)$, so $f(10+x)=f(x)$, which means $f(x)$ is a periodic function with a period $T=10$. Since $f(-3)=f(7) \neq 0$ and $f(3)=0$, $f(x)$ is neither an odd nor an even function.
(2) Because $f(x)$ has only $f(1)=f(3)=0$ in the interval $[0,7]$, and the period $T=10$, the equation $f(x)=0$ has 2 roots in each of the intervals $[0,10), [10,20), [20,30), \cdots, [2000,2005]$, totaling 402 roots. It also has 2 roots in each of the intervals $[-10,0), [-20,-10), \cdots, [-1990,-2000)$. Since $f(-2007)=f(3)=0$ and $f(-2009)=f(1)=0$, the equation $f(x)=0$ has no solutions in the interval $[-2000,-2005]$, thus there are 400 roots.
In summary, the equation $f(x)=0$ has 802 real roots in the interval $[-2005,2005]$.
Note: This is a well-designed problem that is both reasonable and unexpected. Normally, problem (1) should be designed to "prove that the function $f(x)$ is a periodic function," forming a "serial relationship" with problem (2). The brilliance of this problem lies in defying this "norm" by designing it to "determine the odd or even nature of $f(x)$," and the conclusion is unexpectedly that $f(x)$ is neither odd nor even.