Segmentation and paging
Answer to question 1:
If $2^n$ is the size of the page then there are $2^{13-n}$ pages.
We know that each page table entry is 2 bytes.
So size of the page table will be $2\times 2^{13-n}$ bytes because $2^{13-n}$ is the number of pages and each page has an entry in the page table.
They have given that page table takes at most one page to store.
So, $2\times 2^{13-n}=2^n \Rightarrow n=7$.

Answer to question 2:
Each page table entry is two bytes which is $16$ bits so the remaining bits other than the valid bits (protection bits and the dirty bit) are $16-5=11$ bits.
Page size is $512=2^9$ bytes.
Physical address space $2^{16}$ bytes.
Number of bits required to represent a page is $16-9=7$ bits.
So, the available bits to store the aging information is $11-7=4$ bits.