1. Given that $f(x)$ is a function defined on the set of real numbers, and $f(x+2)[1-f(x)]=1+$ $f(x)$. If $f(1)=2+\sqrt{3}$, then the value of $f(1949)$ is $\qquad$ .
1. $\sqrt{3}-2$.
$f(x+2)=\frac{1+f(x)}{1-f(x)}$, so $f(x+4)=\frac{1+f(x+2)}{1-f(x+2)}=-\frac{1}{f(x)}$, then $f(x+8)=$ $-\frac{1}{f(x+4)}=f(x)$, so $f(x)$ is a periodic function with a period of 8. $f(1949)=f(243 \times 8+5)=$ $f(5)=f(1+4)=-\frac{1}{f(1)}=\sqrt{3}-2$.