The internal energy change for a reversible cycle is zero? The internal energy change for a reversible "cycle" is larger than zero? (not reversible "process")
AND, internal energy change for a irreversible cycle is zero?
My book (the book is designed to prepare to pass the national qualification test in Korea—Engineer General Machinery; similar to FE Mechanical in USA. maybe) says that "reversible cycle: $dU>0$, irreversible cycle: $dU=0$"
But, I think it is incorrect.
I can't find errata of the book.
 A: 
My book says that "reversible cycle : dU>0, irreversible cycle : dU=0"

The statement makes no sense. Cycle reversibility only impacts entropy, and only impacts total entropy of a cycle (system + surroundings).
The change in internal energy of the system for any thermodynamic cycle is always zero, regardless of whether the cycle is reversible or not. The change in total energy (internal plus external forms) of the surroundings is also zero since the work done on the surroundings by the system equals the heat transfer to the system from the reservoirs in the surroundings.
The change in entropy, $\Delta S$, of the system for any thermodynamic cycle is also zero, regardless of whether the cycle is reversible or not. That's because entropy, like internal energy is a state property. However, for an irreversible cycle there is an increase in entropy of the surroundings making the total entropy change (system plus surroundings) larger for an irreversible cycle than a reversible cycle. That's because, unlike internal energy, entropy is not a conserved quantity. Entropy is generated ("created", if you will) in an irreversible cycle.
So you can say, for a reversible cycle $\Delta S_{tot}=0$ and for an irreversible cycle $\Delta S_{tot}\gt0$.
Hope this helps.
A: This is for the benefit of @Bob D.
We have already considered the response of the surroundings to the net heat flow from the surroundings to the system $Q_{net}$; this results in a decrease in the internal energy of the surroundings/reservoirs $\Delta U_R$ :  $$\Delta U_R=-Q_{net}$$It is also important to consider the response of the surroundings to  the work W done by the system on the surroundings.  Assuming that the work done on the surroundings results in no additional heat exchange with the reservoirs, we have that $$\Delta U_O+\Delta (PE)+\Delta (KE)=W$$where $\Delta U_O$ is any additional change in internal energy of the surroundings associated with the work that the system does on the surroundings, $\Delta (PE)$ is the change in potential energy of the surroundings, and $\Delta (KE)$ is the change in kinetic energy of the surroundings.  So we have for the overall surroundings that $$\Delta U_S+\Delta (PE)+\Delta (KE)=0$$where $\Delta U_S=\Delta U_R+\Delta U_0$ is the overall change in internal energy of the surroundings.  So, if the work done by the system on the surroundings is fairly useful work, and results in raising a weight and/or accelerating a weight (such as a train, for example), the internal energy of the surroundings decreases.
