Ok, 10 h is slightly simpler.
Yes!
No, you did the 10 h compensation twice.
The 10 h output rate is 0.1, so a .419 m3 battery would deliver 0.419 × 0.8 × 0.1 = 0.0335 MW.
To get the size of the needed battery you have to reverse that calculation: You need 0.335 MW for 10 h, each m3 delivers 0.8 MW for 1 h or 0.8 × 0.1 = 0.08 MW for 10 h, so you need 0.335 / 0.8 / 0.1 = 4.19 m3 of battery.
0.419 × 2 = 0.838 tonnes.
The battery is calculated for full throttle, you can probably get by with a smaller battery, but still very large.
A fuel cell is probably a better alternative.
A TL-10 fuel cell has an output of 0.5 MW for each m3.
You need 0.335 MW, so 0.335 / 0.5 = 0.67 m3 with a mass of 0.67 tonnes and a cost of MCr 0.013.
Fuel consumption is 0.3 kl/h = 0.3 m3/h for each MW, so 0.3 × 0.335 MW × 10 h = 1 m3 for 10 h.