I'll assume that you mean inversions, not transpositions, as bgins suggested. I'll also assume that the question does refer to the number and not the proportion of inversions, since $k$ looks like an integer variable, and that you just confused the two when you said that the expectation value is $1/2$. Also, note that the right-hand side of your inequality is the one for the two-sided Chebyshev inequality, whereas you want the one-sided one.
Finding the variance works similarly to finding the expectation value; it's just a bit more involved. Both use linearity of expectation.
Since every pair of elements has probability $1/2$ of being inverted, the expectation value for the number of inversions is half the number of unordered pairs, $n(n-1)/4$.
For the variance, we need to consider all possible ordered pairs of unordered pairs of distinct elements. There are four kinds of these: $n(n-1)/2$ pairs of identical pairs, $n(n-1)(n-2)/3$ pairs with an inner element in common, $2n(n-1)(n-2)/3$ pairs with an outer element in common, and $n(n-1)(n-2)(n-3)/4$ pairs with no elements in common, for a total of $n(n-1)(1/2+(n-2)/3+2(n-2)/3+(n-2)(n-3)/4)=(n(n-1)/2)^2$ pairs. We need to find the expectation value of the product of the inversion indicator variables for each type of pair.
For the pairs of identical pairs, the product is a square, and the square of an indicator variable is just the variable itself, so this expectation value is also $1/2$.
For the pairs with an inner element in common, all $6$ orders of the three elements are equiprobable and there is only one order in which both pairs are inverted, so the expectation value is $1/6$.
For the pairs with an outer element in common, all $6$ orders of the three elements are equiprobable and there are two orders in which both pairs are inverted, so the expectation value is $1/3$.
For the pairs with no elements in common, the two pairs have an independent probability of $1/2$ each of being inverted, so the expectation value is $1/4$. In total, the expectation value of the square of the number of inversions is
$\frac12\frac{n(n-1)}2+\frac16\frac{n(n-1)(n-2)}3+\frac13\frac{2n(n-1)(n-2)}3+\frac14\frac{n(n-1)(n-2)(n-3)}4\\=\frac{n(n-1)(9n^2-5n+10)}{144}\;.$
Subtracting the square of the expectation value of the number of inversions yields the variance
$\operatorname{Var}T=\frac{n(n-1)(9n^2-5n+10)}{144}-\left(\frac{n(n-1)}4\right)^2=\frac{n(n-1)(2n+5)}{72}\;.$