There is alcohol in 50g alcohol solution A and 350g alcohol solution B: (2000+50+350)×65%= 1560 (g).

Increase alcohol: 1560- 1500=60 (g)

When the concentration ratio of the two alcohol solutions is 3: 1, 50g alcohol solution a is replaced by alcohol solution b, which is equivalent to 50÷ 1/3= 150 (g) alcohol solution b.

The added 60 grams of alcohol is equivalent to 150+350=500 grams of alcohol in alcohol solution b, so the concentration of alcohol solution b is 60÷500= 12%.

Then the concentration of alcohol solution A is 12%×3=36%.

2. The mixed alcohol solution contains pure alcohol: 200×2 1%=42 (liter).

20 liters of ethanol solution contains pure alcohol: 20×20%=4 liters.

200-20= 180 (L) At this time, the volumes of the two alcohol solutions are equal, and the mixed concentration is: (20%+50%)÷2=35%.

If 180L is a mixed alcohol solution of ethylene and propylene, there should be pure alcohol: 180X 35% = 63 (L), which is much more than the actual situation: 63-(42-4)=25 (L).

In fact, some are methanol solutions. When the concentration is 35%, 35%- 10%=25% per liter. This part of the solution is: 25÷25%= 100 (liter).

Then the alcohol solution of B is: (200- 100+20)÷2=60 (liter).

The alcohol solution of C is 60-20=40 (liter).