Given,

$f\left(x\right)+f\left(\frac{1}{1-x}\right)=1+x$

$x=2\Rightarrow f\left(2\right)+f\left(-1\right)=3 \dots \left(1\right)$

$x=-1\Rightarrow f\left(-1\right)+f\left(\frac{1}{2}\right)=0 \dots \left(2\right)$

$x=\frac{1}{2}\Rightarrow f\left(\frac{1}{2}\right)+f\left(2\right)=\frac{3}{2} \dots \left(3\right)$

On subtracting equation $\left(3\right)$ and equation $\left(2\right)$ we get,

$f\left(2\right)-f(-1)=\frac{3}{4} ........\left(4\right)$

On adding equation $\left(1\right)$ and equation $\left(4\right)$ we get,

$2f\left(2\right)=\frac{9}{2}$

$\Rightarrow f\left(2\right)=\frac{9}{4}$