If the mass of the atmosphere is given as 5.1480 × 10^18 kg, then according to the rules of significant figures, the uncertainty in that need not be smaller than 10^14 kg (and depending on how one interprets significant digits, it can be as high as 10^15 kg). According to this site:
And [the exosphere's] mass is only 0.002% of the total mass of the atmosphere because gas molecules are far apart in the exosphere.
That would make it 10^14 kg, within the error bounds allowed by the significant digits, and any difference based on where the exosphere is considered to end would be much smaller.
Also, if we mention the mass of a celestial body, does this value (e.g. in case of Venus 4.867 × 10^24 kg) include its atmosphere's mass or not?
The main way we estimate a planet's mass (and the main reason we care) is its gravitational effects, and apart from probes that have entered the atmosphere, the atmosphere has just as much gravitational effect (per kg) as any other part of the planet.
However, if the mass of Venus is given as 4.867 × 10^24 kg, that implies a error bar no smaller than 10^21 kg.
Wikipedia gives the mass of Venus's atmosphere as 4.8 x 10^20 kg. It also says this is nearly 100 times the mass of Earth's, so Earth's atmosphere would be an even smaller percentage of its total mass.