1) A quadratic can solve this part:
f(x)=Ax2+ Bx + C
f'(x)=2Ax + B
f"(x)=2A
therefore 2A = 4 => A=2
4x+B=0 when x = -1 => B=4
and 2x2 + 4x + C = 2 when x=-1 => C=4
2) A saddle point usually requires a function of two variables...so I am not sure how to answer this part
3) Usually when the 2nd derivative is zero at a critical point there is a horizontal tangent...so f(x)=(x+1)3 has f'(-1)=0 and f"(-1)=0...but x=-1 is NOT a local minimum.....but f(x)=(x+1)4 should serve your purpose