5. Given $f(x)$ 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 $\qquad$
$$
\begin{array}{l}
\text { 5. } \sqrt{3}-2 \text {. } \\
\text { From } f(x+2)[1-f(x)]=1+f(x) \text {, we get } f(x+2) \\
=\frac{1+f(x)}{1-f(x)} \text {. } \\
\therefore \quad f(x+4)=\frac{1+f(x+2)}{1-f(x+2)}=\frac{1+\frac{1+f(x)}{1-f(x)}}{1-\frac{1+f(x)}{1-f(x)}} \\
=-\frac{1}{f(x)} \text {. } \\
\therefore \quad f(x+8)=-\frac{1}{f(x+4)}=f(x) \text {. } \\
\end{array}
$$

Therefore, $f(x)$ is a periodic function with a period of 8.
$$
\begin{aligned}
\therefore \quad & f(1949)=f(243 \times 8+5)=f(5) \\
& =f(1+4)=-\frac{1}{f(1)} \\
& =-\frac{1}{2+\sqrt{3}}=\sqrt{3}-2 .
\end{aligned}
$$