Chemistry - Calculating entropy change: reversible vs irreversible process
You are confused about how to calculate the entropy change for a system and surroundings that have experienced an irreversible process. For the irreversible path, you correctly calculated the final state using the first law of thermodynamics. From this point on, you must totally forget about the irreversible path and, instead, focus exclusively on the initial and final end states. You must separate the system from the surroundings, and then devise an alternative reversible path for each of them separately that takes each from its initial state to its final state. You did that correctly for the system, and determined its correct entropy change (-7.62 J/K). This is the entropy change for the system both for the reversible path you devised as well as for the actual irreversible path.
In the case of the surroundings, during the irreversible process, its internal energy increased by 3.741 kJ at the constant temperature of 300 K. So, for its alternative reversible path, to achieve this same internal energy change at 300 K, the amount of reversible heat it would have to absorb would be the same 3.741 kJ. So, the change in entropy of the surroundings in the irreversible process was 12.47 J/K.
I know that this might all sound a little confusing. So here is a more detailed explanation of the fundamentals as well as a cookbook (foolproof) recipe for determining the change in entropy of the system and surroundings for any arbitrary irreversible process on a closed system (including worked examples of the methodology): https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/