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.
Here is an argument completely different from joriki’s; it is also a complete solution.
A cycle is an odd permutation iff its length is even, so a permutation written as a product of disjoint cycles is even iff an even number of the factors are even cycles. Let $\pi=\sigma_1\dots\sigma_k$ be a permutation of $[n]$ written as a product of $k$ disjoint cycles. Then $n=|\,\sigma_1|+\dots+|\,\sigma_k|$, so the parity of $n$ is the same as the parity of the number of cycles of odd length. Thus, $\pi$ has an even number of cycles of even length iff $n$ and $k$ have the same parity, i.e., iff $(-1)^{n+k}=1$.
Let $e_n$ be the total number of cycles in even permutations of $[n]$, let $o_n$ be the total number of cycles in odd permutations of $[n]$, and let $d_n=e_n-o_n$. The total number of permutations of $[n]$ with $k$ cycles is given by $\left[n\atop k\right]$, the unsigned Stirling number of the first kind. Each of these $\left[n\atop k\right]$ permutations contributes $k$ cycles to $e_n$ if $(-1)^{n+k}=1$, and to $o_n$ if $(-1)^{n+k}=(-1)$. Thus, each contributes $(-1)^{n-k}k$ to $d_n$, and it follows that
$$d_n=\sum_k(-1)^{n+k}k\left[n\atop k\right]\;.$$
Now $(-1)^{n+k}\left[n\atop k\right]=(-1)^{n-k}\left[n\atop k\right]$ is the signed Stirling number of the first kind, for which we have the generating function $$\sum_k(-1)^{n+k}\left[n\atop k\right]x^k=x^{\underline{n}}\;.\tag{1}$$
(Here $x^{\underline{n}}=x(x-1)(x-2)\cdots(x-n+1)$ is the falling factorial, sometimes written $(x)_n$.)
Differentiate $(1)$ with respect to $x$ to obtain
$$\sum_k(-1)^{n+k}k\left[n\atop k\right]x^{k-1}=Dx^{\underline{n}}=(x-n+1)Dx^{\underline{n-1}}+x^{\underline{n-1}}\;,$$
where the last step is simply the product rule, since $x^{\underline{n}}=x^{\underline{n-1}}(x-n+1)$. But $$Dx^{\underline{n-1}}=\sum_k(-1)^{n-1+k}k\left[{n-1}\atop k\right]x^{k-1}\;,$$ so
$$\sum_k(-1)^{n+k}k\left[n\atop k\right]x^{k-1}=\sum_k(-1)^{n-1+k}k\left[{n-1}\atop k\right]x^{k-1}+x^{\underline{n-1}}\;.\tag{2}$$
Now evaluate $(2)$ at $x=1$ to get $$d_n=(2-n)d_{n-1}+[n=1]\;,\tag{3}$$ where the last term is an Iverson bracket. If we set $d_0=0$, $(3)$ yields $d_1=[1=1]=1$, which by direct calculation is the correct value: the only permutation of $[1]$ is the identity, which is even and has one cycle. It’s now a trivial induction to check that $d_n=(-1)^n(n-2)!$ for all $n\ge 1$: the induction step is
$$\begin{align*}d_{n+1}&=\Big(2-(n+1)\Big)d_n\\
&=(1-n)(-1)^n(n-2)!\\
&=(-1)^{n+1}(n-1)!\;.
\end{align*}$$
Added: It occurs to me that there’s a rather easy argument that does not use generating functions. Let $\sigma$ be any $k$-cycle formed from elements of $[n]$; then $\sigma$ is a factor in $(n-k)!$ permutations of $[n]$. Moreover, exactly half of these permutations are even unless $n-k$ is $0$ or $1$. Thus, $\sigma$ contributes to $d_n$ iff $k=n$ or $k=n-1$.
There are $(n-1)!$ $n$-cycles; they are even permutations iff $n$ is odd, so they contribute $(-1)^{n-1}(n-1)!$ to $d_n$.
There are $n(n-2)!$ $(n-1)$-cycles: there are $n$ ways to choose the element of $n$ that is not part of the $(n-1)$-cycle, and the other $n-1$ elements can be arranged in $(n-2)!$ distinct $(n-1)$-cycles. The resulting permutation of $[n]$ is even iff $n-1$ is odd, i.e., iff $n$ is even, so they contribute $(-1)^nn(n-2)!$ to $d_n$.
It follows that $$\begin{align*}d_n&=(-1)^nn(n-2)!+(-1)^{n-1}(n-1)!\\
&=(-1)^n\Big(n(n-2)!-(n-1)!\Big)\\
&=(-1)^n(n-2)!\Big(n-(n-1)\Big)\\
&=(-1)^n(n-2)!\;.
\end{align*}$$
Best Answer
It is a matter of cycle notation.
$\{1, 2, 3, 4\}$ is the set. $(1\; 2\; 3\; 4)$ is a permutation of the set; a sequence of transpositions of that set represented by the rotation of elements. (A transposition is a permutation resulting from an exchange of just two elements.)
This permutation can also be expressed as: $\begin{pmatrix}1 & 2& 3 &4\\ 2 & 3 & 4 & 1\end{pmatrix}$ (Cauchy notation; initial state above, end state below).
It means that the first element is exchanged with the second, then the third, and finally the fourth. This is also expressed as a product of the individual transpositions: $(1\; 2\; 3\; 4) = (1\; 2)(1\; 3)(1\; 4)$ That is an odd number of transpositions (three), meaning it has an odd parity.
$(1 \; 2)$ is a single transposition; so it's an odd parity. $(1\; 2\; 4)$ is two transpositions; so it's an even parity.
$(1\; 2) \equiv \begin{pmatrix}1 & 2 & 3 & 4 \\ 2 & 1 & 3 & 4\end{pmatrix}$
$(1\; 2\; 3) = (1\; 2)(1\; 3) \equiv \begin{pmatrix}1 & 2 & 3 & 4 \\ 2 & 3 & 1 & 4\end{pmatrix}$