Question 1182951
Will assume that {{{f(x) = 2/(ax+b)}}}

f(0) = -2 ==> f(0) = 2/b = -2  ==>  b = -1  ==>   {{{f(x) = 2/(ax-1)}}}

f(2) = 2  ==>  2/(2a - 1) = 2  ==> a = 1  ==>  {{{f(x) = 2/(x-1)}}}

f(x) = x ==>  {{{2/(x-1) = x}}}  ==> 2 = x(x-1)  ==> {{{x^2-x-2=0}}}

<==> (x-2)(x+1) = 0  ==> x = 2 or -1.

Finally, {{{f(p) + f(-p) =  2/(p-1) + 2/(-p-1) = 2*((-2)/(1-p^2)) = 2*(2/(p^2-1)) = 2f(p^2)}}}.