jmc

We know that $f(0)=0,$ so from property (iii), $$f(1)=1-f(0)=1.$$Then from property (iv), $$f\left(\frac{1}{3}\right)=\frac{f(1)}{2}=\frac{1}{2}.$$Then from property (iii), $$f\left(\frac{2}{3}\right)=1-f\left(\frac{1}{3}\right)=1-\frac{1}{2}=\frac{1}{2}.$$Property (ii) states that the function is non-decreasing. Since $f\left(\frac{1}{3}\right)=f\left(\frac{2}{3}\right)=\frac{1}{2},$ we can say that $f(x)=\frac{1}{2}$ for all $\frac{1}{3}\le x\le \frac{2}{3}.$ In particular, $f\left(\frac{3}{7}\right)=\frac{1}{2}.$

Then by property (iv), $$f\left(\frac{1}{7}\right)=\frac{f(\frac{3}{7})}{2}=\frac{1}{4}.$$By property (iii), $$f\left(\frac{6}{7}\right)=1-f\left(\frac{1}{7}\right)=1-\frac{1}{4}=\frac{3}{4}.$$Finally, by property (iv), $$f\left(\frac{2}{7}\right)=\frac{f(\frac{6}{7})}{2}=\boxed{\frac{3}{8}}.$$The properties listed in the problem uniquely determine the function $f(x).$ Its graph is shown below:



For reference, the function $f(x)$ is called the Cantor function. It is also known as the Devil's Staircase.