Using the provided data, estimate the quantity of lubricant required in 2025 when machines are planned to work for 8 x 105 hours. Calculate the slope (b) and y-intercept (a) for the linear regression

The regression equation is $Y = 3.4497 + 1.8041X$.
The total machine hours for 2025 are $8 \times 10^5$ hours. Since $X$ is in $10^4$ hours, we convert:

Since $Y$ is in $10^3$ gallons, the quantity of lubricant is $147.7777 \times 10^3$ gallons.
The estimated quantity of lubricant required in Year 2025 is $\boxed{\text{147.78} \times 10^3 \text{ gallons}}$.

The lubricating oil consumption is $15 \times 10^3$ gallons, so $Y = 15$ in our units.

Since $X$ is in $10^4$ hrs, the total production hour is $6.4022 \times 10^4$ hrs.
The expected total production hour is $\boxed{\text{6.40} \times 10^4 \text{ hours}}$.

From part (i), the total lubricant required in 2025 is $147.7777 \times 10^3$ gallons, which is 147,777.7 gallons.

Total working days per annum = 250 days
Hours per day = 8 hours
Total hours per Oiler per annum = $250 \text{ days/year} \times 8 \text{ hours/day} = 2000 \text{ hours/year}$.

Rate of work for one Oiler = 100 gallons/hour.
Factor for Oiler = 0.8.
Effective rate of work = $100 \text{ gallons/hour} \times 0.8 = 80 \text{ gallons/hour}$.
Effective gallons an Oiler can handle per annum = $2000 \text{ hours/year} \times 80 \text{ gallons/hour} = 160,000 \text{ gallons/year}$.

Since you cannot have a fraction of an Oiler, we round up to the nearest whole number.
The number of oilers required in year 2025 is $\boxed{\text{1 Oiler}}$.