(1)
f(x)=(x^2+a)／(x+1)
f'(x) = [(x^2+a)- 2(x+1)x ]/ (x+1)^2
f'(1) = (1+a -4)/4 = 1/2
-3+a = 2
a = 5
(2)
f'(1) = 0
=> (1+a -4)/4 = 0
a = 3
f'(x) = [(x^2+3)- 2(x+1)x ]/ (x+1)^2
= [-x^2-2x+3]/(x+1)^2
put f'(x) = 0
-x^2-2x+3 =0
-(x+3)(x-1) = 0
x = 1 or -3
f''(x) =[(-x^2-2x+3)2(x+1) - (x+1)^2(-2x-2)]/(x+1)^4
f''(1) = [0- 4(-4)]/16 = 1 >0 (min)
f''(-3)