(1)令x=y=1,f(2)=f(1)+f(1)+12+2k+3⇒k=0,则f(x+y)=f(x)+f(y)+3xy(x+y+2)+3 对于x,y∈R都成立 令x=t(t∈N*),y=1f(t+1)=f(t)+f(1)+3t(t+3)+3⇒f(t+1)-f(t)=3t2+9t+4⇒f(2)-f(1)=3×12+9×1+4f(3)-f(2)=3×22+9×2+4 …f(t)=f(t-1)=3(t-1)2+9×(t-1)+4 用叠加法 f(t)-f(1)=3[12+22+…+(t-1)2]+9[1+2+…+(t-1)]+4+4•k…+4=3•(t-1)•t•(2t-1)+9•(t-1)•t+4(t-1)=t3+3t2-4 ∴f(t)=t3+3t2-3(t≥2)又t=1适合上式f(t)=t3+3t2-3(t≥2)(t∈N*) (2)若t∈N*,则t3+3t2-3=t⇒t3-t+3(t2+1)=0⇒(t-1)(t+1)(t+3)=0⇒t1=1,t2=-1,t3=-3(舍) 又令x=y=0f(0)=-3 令y=-x-3=f(x)+f(-x)+(-6x2)+3⇒f(x)+f(-x)=6x2-6对x∈R都成立 若t为负整数,则f(t)=6t2-6-f(-t)=6t2-6+t3-3t2+3=t3+3t2-3 由t3+3t2-3=t(t+3)(t+1)(t-1)⇒t1=-3,t2=-1,t3=1(舍) 若t=0,则f(t)═t无解 综上,满足f(t)=t,所有整数t为1,-1,-3; (3)要使不等式恒成立,则只需m≤对t≥4,且t∈N*恒成立 即m≤(t+3)(t+1)(t-1) | (t+3)(t+1) | =t-1对t≥4,且t∈N*恒成立 即且m≤(t-1)min=3 实数m最大值为3 |