A hospital uses a mechanical lift device to raise 400 kg patient support platform through a vertical height of 1.5m.

Given: V.R. = 8, Load Distance = 0.5 m

$$\text{Effort Distance} = \text{V.R.} \times \text{Load Distance}$$ $$\text{Effort Distance} = 8 \times 0.5 \, \text{m}$$ $$\text{Effort Distance} = 4 \, \text{m}$$

The distance moved by the effort is $\boxed{4 \, \text{m}}$.

Given: Load = 2600 N, Load Distance = 0.5 m

$$\text{Output Work} = 2600 \, \text{N} \times 0.5 \, \text{m}$$ $$\text{Output Work} = 1300 \, \text{J}$$

The output work is $\boxed{1300 \, \text{J}}$.

Given: Effort = 500 N, Effort Distance = 4 m

$$\text{Input Work} = 500 \, \text{N} \times 4 \, \text{m}$$ $$\text{Input Work} = 2000 \, \text{J}$$

The input work is $\boxed{2000 \, \text{J}}$.

Given: Input Work = 2000 J, Output Work = 1300 J

$$\text{Energy Lost} = 2000 \, \text{J} - 1300 \, \text{J}$$ $$\text{Energy Lost} = 700 \, \text{J}$$

The energy lost is $\boxed{700 \, \text{J}}$.

Given: Input Work = 2000 J, Time = 60 s

$$\text{Input Power} = \frac{2000 \, \text{J}}{60 \, \text{s}}$$ $$\text{Input Power} \approx 33.33 \, \text{W}$$

The input power is $\boxed{33.33 \, \text{W}}$.

Useful Output Power:

Given: Output Work = 1300 J, Time = 60 s

$$\text{Useful Output Power} = \frac{1300 \, \text{J}}{60 \, \text{s}}$$ $$\text{Useful Output Power} \approx 21.67 \, \text{W}$$

The useful output power is $\boxed{21.67 \, \text{W}}$.

The problem description for the hospital lift does not contain a specific question to answer.