To complete the table, we need to calculate the interest and future value for each year of the investment. The formula for the future value ( A ) of an investment with principal ( P ), annual interest rate ( r ), and time ( t ) in years is given by:

[ A = P(1 + r)^t ]

The interest earned ( I ) is the difference between the future value and the principal:

[ I = A - P ]

Let's break it down step by step.

Year 1:

1. Principal amount (P): P800,000
2. Interest rate (r): 7.5% = 0.075
3. Time (t): 1 year

Using the future value formula:

[ A_1 = 800,000(1 + 0.075)^1 = 800,000 \times 1.075 = 860,000 ]

The interest earned in the first year is:

[ I_1 = A_1 - P = 860,000 - 800,000 = 60,000 ]

So, the completed row for Year 1 is:

[ \begin{array}{|c|c|l|l|c|l|} \hline \text{Time} & \text{Principal amount} & \text{Interest rate} & \text{Solution} & \text{Interest} & \text{Future value} \ \hline 1 & P800,000 & 7.5% & (P800,000)(1 + 0.075) & P60,000 & P860,000 \ \hline \end{array} ]

Year 2:

1. Principal amount (P): P860,000 (the future value from Year 1)
2. Interest rate (r): 7.5% = 0.075
3. Time (t): 1 year

Using the future value formula:

[ A_2 = 860,000(1 + 0.075)^1 = 860,000 \times 1.075 = 924,500 ]

The interest earned in the second year is:

[ I_2 = A_2 - P = 924,500 - 860,000 = 64,500 ]

So, the completed row for Year 2 is:

[ \begin{array}{|c|c|l|l|c|l|} \hline \text{Time} & \text{Principal amount} & \text{Interest rate} & \text{Solution} & \text{Interest} & \text{Future value} \ \hline 2 & P860,000 & 7.5% & (P860,000)(1 + 0.075) & P64,500 & P924,500 \ \hline \end{array} ]

Combining both years, the completed table is:

[ \begin{array}{|c|c|l|l|c|l|} \hline \text{Time} & \text{Principal amount} & \text{Interest rate} & \text{Solution} & \text{Interest} & \text{Future value} \ \hline 1 & P800,000 & 7.5% & (P800,000)(1 + 0.075) & P60,000 & P860,000 \ \hline 2 & P860,000 & 7.5% & (P860,000)(1 + 0.075) & P64,500 & P924,500 \ \hline \end{array} ]

The final answer is:

[ \boxed{P924,500} ]