According to the meaning of the question, each bundle of A, B and C is 1500 yuan, 2000 yuan and 3000 yuan respectively. Because only two kinds of lottery tickets are needed, it is good to combine them in pairs. Let the number of bundles bought by A, B and C be X, Y and Z respectively.
When buying A and B, X+Y=20, 1500X+2000Y=45000, the solution is X=- 10, Y=30, and x cannot be negative, so this scheme is not feasible.
2. When buying B and C, there are Y+Z=20, 2000Y+2500Z=45000, Y= 10 and Z= 10, so this scheme is feasible, and B and C buy 10 packages respectively;
3. If you buy A and C, there are X+Z=20, 1500X+2500Z=45000, and the solution is X=5, Z= 15, so this scheme is feasible. A and C buy 5 bundles and 15 bundles respectively;
The second question: Just compare the two kinds of handling fees.
When buying B and C lottery tickets,10 * 0.3 *1000+10 * 0.5 *1000 = 8000 yuan.
When buying lottery tickets A and C, there are 5 * 0.2 *1000+15 * 0.5 *1000 = 8500 yuan.
Obviously, 8500 is more than 8000, so choose A and C to buy 5 bundles and 15 bundles respectively.