To find out how much water the tank holds two hours later, we need to integrate the rate function $r(t) = 5 - 5e^{-0.2t}$ from $t = 0$ to $t = 120$ (since two hours is 120 minutes).

Step 1: Integrate the rate function

$\int_{0}^{120} (5 - 5e^{-0.2t}) \, dt$

This can be split into two separate integrals:

$\int_{0}^{120} 5 \, dt - \int_{0}^{120} 5e^{-0.2t} \, dt$

The first integral:

$\int 5 \, dt = 5t$

Evaluated from 0 to 120:

$5(120) - 5(0) = 600$

The second integral:

$-5 \int e^{-0.2t} \, dt = -5 \left( -\frac{1}{0.2} e^{-0.2t} \right) = 25e^{-0.2t}$

Evaluated from 0 to 120:

$25(e^{-0.2 \times 120} - e^{0}) = 25(0 - 1) = -25 \cdot (-1) = 25(1 - e^{-24})$

Since $e^{-24}$ is extremely small, $25(1 - e^{-24}) \approx 25$.

Step 2: Subtract the integrated values from the initial amount

The total water pumped out is $600 - 25 = 575$ liters.

The tank originally had 800 liters, so the amount remaining is:

$800 - 575 = 225 \, \text{liters}$

Therefore, the tank holds approximately 225 liters of water two hours after the pump starts. The answer is:

225 Liters