(a,b,c) When ${x_1} = – 1$ and ${x_2} = 1$ , then

$f( – 1) – f(1) = f\left[ {\frac{{ – 1 – 1}}{{1 + 1(1)}}} \right] = f( – 1) \Rightarrow f(1) = 0$

Which is satisfied when $f(x) = {\tan ^{ – 1}}\left( {\frac{{1 – x}}{{1 + x}}} \right)$

When ${x_1} = {x_2} = 0$, then

$f(0) – f(0) = f\left[ {\frac{{0 – 0}}{{1 – 0}}} \right] = f(0) \Rightarrow f(0) = 0$

When ${x_1} = – 1$ and ${x_2} = 0$ then

$f( – 1) – f(0) = f\left( {\frac{{ – 1 – 0}}{{1 – 0}}} \right) = f( – 1) \Rightarrow f(0) = 0$

Which is satisfied when $f(x) = \log \left( {\frac{{1 – x}}{{1 + x}}} \right)$ and 

$f(x) = \log \left( {\frac{{1 + x}}{{1 – x}}} \right)$.