I do the rest for my too much mistakes.

2)In fact, we can begain at any point,...n=k OK, see n=k+1,we can take k points, if one point out,...but it beetween i and i+1 steps,...,it should be in , contradiction....

3) f(0)=f(1+(-1))<=(-1)f(1)+f(f(1))=-f(1+0)+f(f(1))<=-[0*f(1)+f(f(1))]+f(f(1))=0

   f(x)=f((x+1)+(-1)<=-f(x+1)+f(f(x+1))=-[f((x+1)+0)]+f(f(x+1))<=-[0*f((x+1))+f(f(x+1))]+f(f(x=1))=0, So,f(x)=0

The condition is too strong,only f(x)=0. this Q,first line is not necessary.

5)[f(2)-f(1)]/f(1)=k2,f(2)=(k2+1)f(1). [f(3)-f(2)]/f(1)=k3,...f(3)=(k3+k2+1)f(1),..

f(i)=(ki+k(i-1)+...k2+1)f(1).,

[f(n)-f(n-m)]/f(m)=K. [(kn+...+k2+1)f(1)-(k(m-n)+...+k2+1)f(1)]/(km+...+k2+1)f(1)=K

[kn+...+k(m+1)]/(km+...+k2+1)=K

f(n)/f(m)=[(kn+...+k2+1)f(1)/(km+...k2+1)f(1)=[kn+...+k(m+1)]/(km+...k2+1)=K+1.OK.

By the way.{the condition also strong,if f(0)is small,there are a lot f(s)=f(t)}

6)Too simple to try.