Given, $f\left(x\right)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right), -1<x<1$ and $g\left(x\right)=\sqrt{3+4x-4x^2}$

Since, we are given with the domain of $f\left(x\right),$ then the domain $\left(f+g\right)$ can be obtained by finding the domain of $g\left(x\right)$ and taking the intersection with the domain of $f\left(x\right).$

Now, for $g\left(x\right)$ to be defined $3+4x-4x^2\ge 0$

$\Rightarrow 4x^2-4x-3\le 0$

$\Rightarrow \left(2x+1\right)\left(2x-3\right)\le 0$

Thus, the domain of $g\left(x\right)$ is $\left[-\frac{1}{2}, \frac{3}{2}\right].$

Now, the domain of $\left(f+g\right)$ is $\left(-1, 1\right)\cap \left[-\frac{1}{2}, \frac{3}{2}\right]\equiv \left[-\frac{1}{2}, 1\right)$