The information specified is insufficient to determine the desired parities.
For each of the parities in question, we can find two examples, each satisfying the given conditions, but where the parities of the results are not the same.
For the question of the parity of $(x^{-1}\;\text{mod}\;p)$, let $p=5$.$\;$Then
For $a=2$, we get $(a^{-1}\;\text{mod}\;p)=3$.
For $a=4$, we get $(a^{-1}\;\text{mod}\;p)=4$.
For $c=1$, we get $(c^{-1}\;\text{mod}\;p)=1$.
For $c=3$, we get $(c^{-1}\;\text{mod}\;p)=2$.
For the question of the parity of $(xy\;\text{mod}\;p)$, let $p=7$.$\;$Then
For $a=2,\,b=4$, we get $(ab\;\text{mod}\;p)=1$.
For $a=4,\,b=4$, we get $(ab\;\text{mod}\;p)=2$.
For $a=4,\,c=3$, we get $(ac\;\text{mod}\;p)=5$.
For $a=4,\,c=5$, we get $(ac\;\text{mod}\;p)=6$.
For $c=1,\,d=1$, we get $(cd\;\text{mod}\;p)=1$.
For $c=3,\,d=3$, we get $(cd\;\text{mod}\;p)=2$.
Thus, as shown by the above examples, the parity, mod $p$, of products and inverses, is not uniquely determined by the parities, mod $p$, of the operands.
Suggestion: In the future, do some initial testing to avoid asking questions about the truth of statements which are so easily proven false.
Consider any listing of the numbers $1, 2,...,2n$ that obey the following two conditions: (1) The even numbers appear in order (not necessarily consecutively), and the last number in the list is $2n$. For example if $n=4$, the following would be a permissible list: $5,1,2,4,3,7,6,8$.
Each such listing corresponds to a permissible permutation of $1, 2,...,2n$, where a cycle ends with an even number. For instance in the example above, the permutation would be $(5,1,2)(4)(3,7,6)(8)$.
And this correspondence is reversible. Given a permissible permutation, we can get such a listing, by just putting the even number in each cycle at the end of the cycle, and then removing the cycle parentheses to get the listing.
Now start with the even numbers in their proper order $2,4,6,...,2n$. Next place $1$. There are $n$ choices for this--$1$ can go before any of the even numbers. Next place $3$. There are $n+1$ choices here ($3$ could go before any number currently in the list, which now includes $1$ and all the evens). Next place $5$ ($n+2$ ways to do this). Etc.
Altogether there are $n\cdot (n+1)\cdot (n+2)\cdot \ldots \cdot (2n-1)$ ways to construct these sequences; and consequently that is the number of admissible permutations as well.
Best Answer
Parity and number of inversions go together: if the number of inversions is even, so is the parity, and if the number of inversions is odd, so is the parity. Thus, both of these boil down to counting inversions. Every time a larger number precedes a smaller number in a permutation, you have an inversion.
Let’s look at your third example, $259148637$. The $2$ precedes $5,9,1,4,8,6,3$, and $7$; the only one of these that is smaller than $2$ is $1$, so the pair $(2,1)$ is the only inversion involving the $2$. Go on to the $5$: it precedes $9,1,4,8,6,3$, and $7$; of these, $1,4$, and $3$ are smaller than $5$, so we get another three inversions: $(5,1),(5,4)$, and $(5,3)$. Continue in the same way: the $9$ is larger than all six of the numbers that follow it, so it contributes six inversions; the $1$ isn’t larger than any of the later numbers, so it contributes none. And so on: the $4$ contributes one, $(4,3)$; the $8$ contributes three, since it’s larger than all three of the later numbers; the $6$ contributes one, $(6,3)$; and that’s it, since the $3$ and $7$ contribute none.. The grand total is therefore $$1+3+6+0+1+3+1+0+0= 15\;,$$ if I’ve not miscounted. We conclude that the permutation $259148637$ has $15$ inversions, and since $15$ is odd, it’s an odd permutation.
Finding the inverse is another matter altogether. For this it’s easiest to write out the permutation in two-line notation, like this: $$\matrix{1&2&3&4&5&6&7&8&9\\2&5&9&1&4&8&6&3&7}\tag{1}$$ This is basically just a tabular display of the permutation as a function, one that takes each number in the top row to the number below it. If I call the permutation $\pi$, I can think of it as the function such that $\pi(1)=2,\pi(2)=5,\pi(3)=9,\dots,\pi(9)=7$. The inverse function just reverses all of these pairs: $\pi^{-1}(2)=1,\pi^{-1}(5)=2,\pi^{-1}(9)=3,\dots,\pi^{-1}(7)=9$. In tabular form this is just turning $(1)$ upside-down: $$\matrix{2&5&9&1&4&8&6&3&7\\1&2&3&4&5&6&7&8&9}\tag{2}$$
Now $(2)$ is a bit hard to use, because the top (or ‘input’) row is out of order. To fix this, just reorder the columns so that the top row is in numerical order: $$\matrix{1&2&3&4&5&6&7&8&9\\4&1&8&5&2&7&9&6&3}\tag{3}$$ To get $(3)$ from $(2)$ I just moved the columns as units: the fourth column of $(2)$ becomes the first column of $(3)$, the first column of $(2)$ becomes the second column of $(3)$, the eighth column of $(2)$ becomes the third column of $(3)$, and so on, all the way down to the third column of $(2)$, which $-$ since it has the $9$ in the top row $-$ becomes the last column of $(3)$.
Now just read off the bottom row of $(3)$: $418527963$ is the inverse of the original permutation. Try this procedure on the others, and see whether you have any questions.