Using first equation we get \[f(m+88) \leq f(m)+88\] and from the second one \[f(m+88) \geq f(m)+88\] which means that in all those inequalities, the equality holds. We got \[f(m+8)=f(m)+8\] and \[f(m+11)=f(m)+11\] Let $f(0)=a$. As $GCD(8,11)=1$. Each number $k$ can be written in form \[k=x \cdot 8+y \cdot 11\] where $x,y \in Z$. Using our equalities we obtain $f(k)=x \cdot 8+y \cdot 11=k$ which means that \[f(k)=k+a\] for some $a \in Z$

Another solution. We prove that the answer is $f(m)=m+a$ for all $m \in \Bbb{Z}$,and some $a \in \Bbb{Z}$. We know that $i)$ $f(m+8) \leq f(m)+8$ $ii)$ $f(m+11) \geq f(m)+11$ By $i$ we get that: $$f(m)+11 \geq f(m+8)+3$$So by $ii$ we get that: $$f(m+11) \geq f(m+8)+3$$ Hence $$f(m) \geq f(m-3)+3 \quad(1)$$ Take a fixed integer $m_{0}$. By using $(1)$ and $i$ we get the bellow result: $$f(m_{0}+2)+6 \leq f(m_{0}+8) \leq f(m_{0})+8$$$$f(m_{0})+9 \leq f(m_{0}+9) \leq f(m_{0}+1)+8$$$$f(m_{0}+1)+9 \leq f(m_{0}+10) \leq f(m_{0}+2)+8$$ So; $$f(m_{0}+2)+6 \leq f(m_{0})+8$$$$f(m_{0})+9 \leq f(m_{0}+1)+8$$$$f(m_{0}+1)+9 \leq f(m_{0}+2)+8$$ As the sum of LHS’s is equal to sum of RHS’s; inequalities become to equality. So: $$f(m_{0}+2)=f(m_{0})+2$$$$f(m_{0})=f(m_{0}+1)-1$$$$f(m_{0}+1)=f(m_{0}+2)-1$$It means that: $$f(m_{0})=f(m_{0}+1)-1=f(m_{0}+2)-2$$ Which holds for any integer $m$. Hence it concludes that $f(m+1)=f(m)+1$ for all integer $m$. So $f(m)=m+a$ for all $m \in \Bbb{Z}$ and some $a \in \Bbb{Z}$. And it holds problem’s inequalities. Done.