After much head banging.
Part 2): 256K(218) of memory. The offset can address 28 bytes. We have to use 218/28 that is 210 pages. Hence the 1024 Address space. What this means is you need 1024 pages to access 256K Memory. Now how many 3rd level pages are required to accomodate 1024 pages ? 2^10/2^6 = 2^4 = 16 Pages.
Therefore we need 16 entries in a 2nd level page table, and one entry in the first level page table
The rest you can follow from the given solution.
Part 3): Ist level page table in 10 bits long. So, 64 K starting at address 0x1000000, 0001 0000 0000 ......
600K starting at address 0x80000000 1000 0000 0000 .....
64K starting at address 0xf0000000 1111 0000 0000 .....
So the first 10 bits tell us that 3 different 1st level pages.
2^64/2^8 = 2^8 pages ----> 2^8/2^6 = 4 pages and so on, the rest you can follow from the given solution.
