Part D:

For the variation of the internal energy in the path cda we have that

\(\displaystyle\triangle{U}_{{{c}{a}}}=\triangle{U}_{{{c}{d}}}+\triangle{U}_{{{d}{a}}}\)

\(\displaystyle\triangle{U}_{{{c}{a}}}={\left({Q}_{{{c}{d}}}-{W}_{{{c}{d}}}\right)}+{\left({Q}_{{{d}{a}}}-{W}_{{{d}{a}}}\right)}\)

the path da is a constant volume, so the work in this path is zero

\(\displaystyle\triangle{U}_{{{c}{a}}}={\left({Q}_{{{c}{d}}}-{W}_{{{c}{d}}}\right)}+{\left({Q}_{{{d}{a}}}-{0}\right)}\)

\(\displaystyle\triangle{U}_{{{c}{a}}}={Q}_{{{c}{d}}}+{Q}_{{{d}{a}}}-{W}_{{{c}{d}}}\)

\(\displaystyle\triangle{U}_{{{c}{a}}}={Q}_{{{c}{a}}}-{W}_{{{c}{d}}}\)

Here we have, form the parts B and C, the \(\displaystyle{Q}_{{{c}{a}}}={130}{J}\) and the \(\displaystyle{W}_{{{c}{d}}}={25}\ {J}\), so

\(\displaystyle\triangle{U}_{{{c}{a}}}={105}{J}\)

An Answer that you already said is incorrect. But there is a "trick", the significant figures. You need an answer with two significant figures but that has three. So, rounding, the answer is

\(\displaystyle\triangle{U}_{{{c}{a}}}={110}{J}\)

Part E:

Since the path dc is at constant volume

\(\displaystyle\triangle{U}_{{{c}{d}}}={Q}_{{{c}{d}}}\)

And like we have \(\displaystyle{Q}_{{{c}{d}{a}}}\)

\(\displaystyle{Q}_{{{c}{d}{a}}}={Q}_{{{c}{d}}}+{Q}_{{{d}{a}}}\)

\(\displaystyle{Q}_{{{d}{a}}}={Q}_{{{c}{d}{a}}}-{Q}_{{{c}{d}}}\)

\(\displaystyle{Q}_{{{d}{a}}}={130}{J}-{42}{J}\)

\(\displaystyle{Q}_{{{d}{a}}}={88}{J}\)

For the variation of the internal energy in the path cda we have that

\(\displaystyle\triangle{U}_{{{c}{a}}}=\triangle{U}_{{{c}{d}}}+\triangle{U}_{{{d}{a}}}\)

\(\displaystyle\triangle{U}_{{{c}{a}}}={\left({Q}_{{{c}{d}}}-{W}_{{{c}{d}}}\right)}+{\left({Q}_{{{d}{a}}}-{W}_{{{d}{a}}}\right)}\)

the path da is a constant volume, so the work in this path is zero

\(\displaystyle\triangle{U}_{{{c}{a}}}={\left({Q}_{{{c}{d}}}-{W}_{{{c}{d}}}\right)}+{\left({Q}_{{{d}{a}}}-{0}\right)}\)

\(\displaystyle\triangle{U}_{{{c}{a}}}={Q}_{{{c}{d}}}+{Q}_{{{d}{a}}}-{W}_{{{c}{d}}}\)

\(\displaystyle\triangle{U}_{{{c}{a}}}={Q}_{{{c}{a}}}-{W}_{{{c}{d}}}\)

Here we have, form the parts B and C, the \(\displaystyle{Q}_{{{c}{a}}}={130}{J}\) and the \(\displaystyle{W}_{{{c}{d}}}={25}\ {J}\), so

\(\displaystyle\triangle{U}_{{{c}{a}}}={105}{J}\)

An Answer that you already said is incorrect. But there is a "trick", the significant figures. You need an answer with two significant figures but that has three. So, rounding, the answer is

\(\displaystyle\triangle{U}_{{{c}{a}}}={110}{J}\)

Part E:

Since the path dc is at constant volume

\(\displaystyle\triangle{U}_{{{c}{d}}}={Q}_{{{c}{d}}}\)

And like we have \(\displaystyle{Q}_{{{c}{d}{a}}}\)

\(\displaystyle{Q}_{{{c}{d}{a}}}={Q}_{{{c}{d}}}+{Q}_{{{d}{a}}}\)

\(\displaystyle{Q}_{{{d}{a}}}={Q}_{{{c}{d}{a}}}-{Q}_{{{c}{d}}}\)

\(\displaystyle{Q}_{{{d}{a}}}={130}{J}-{42}{J}\)

\(\displaystyle{Q}_{{{d}{a}}}={88}{J}\)