Answer
(b) $\frac{3}{2}$
(c) $\frac{\pi}{4}$
(a) See explanation
Work Step by Step
(a) $- d x=du$ and $-x+a=u$
And then $I=-\int_{a}^{0} \frac{(-u+a)f}{(-u+a)f+f(u)} d u=\int_{0}^{a} \frac{(-u+a)f}{(-u+a)f+f(u)} d u=\int_{0}^{a} \frac{f(a-x)}{f(a-x)+f(x)} d x$
$=\int_{0}^{a} \frac{(-x+a)f+f(x)-f(x)}{(-x+a)f+f(x)} d x=\int_{0}^{a} 1 d x-\int_{0}^{a} \frac{f(x)}{(-x+a)f+f(x)} -1+a=d x$
So $I=-I+a \Rightarrow \frac{a}{2}= I$
(b) $3=a,$ so the integral is then $I=\frac{3}{2}$
(c) $\frac{\pi}{2}=a,$ so the integral is then $I=\frac{\pi / 2}{2}=\frac{\pi}{4}$
