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
$$\operatorname{Var}T=\frac{n(n-1)(9n^2-5n+10)}{144}-\left(\frac{n(n-1)}4\right)^2=\frac{n(n-1)(2n+5)}{72}\;.$$
Before the solution, a minor comment.
The Chebyshev Inequality is not quite quoted correctly. It should be
$$\Pr(|X-\mu|\ge k\sigma)\le \frac{1}{k^2}.\tag{$1$}$$
For continuous distributions there is no need to distinguish between $\le$ and $\lt$. Here we are working with a discrete distribution.
A standard calculation shows that in our case $\mu=np=70$ and $\sigma^2=np(1-p)=35$. We want a lower bound on $\Pr(60\lt X\lt 80)$. The complementary event is $|X-70|\ge 10$. We first find an upper bound for $\Pr(|X-70|\ge 10)$.
Compare with Inequality $(1)$ quoted above. In our case we have $k\sigma=10$, and therefore
$$k=\frac{10}{\sigma}, \quad\text{so}\quad\frac{1}{k^2}=\frac{\sigma^2}{100}=\frac{35}{100}.$$
It follows that $\Pr(|X-70|\ge 10)$ is $\le \frac{35}{100}$. Thus
$$\Pr(60\lt X\lt 80)\ge 1-\frac{35}{100}=\frac{65}{100}.$$
That is the lower bound given by the Chebyshev Inequality.
Remark: It is not a very good lower bound. You might want to use software such as the free-to-use Wolfram Alpha to calculate the exact probability. It's not Chebyshev's fault. An inequality that works for every distribution that has a mean and variance, including some pretty weird ones, cannot be expected to compete against estimates based on more information.
Best Answer
$X$ is the result of a fair (unbiased) die roll. The first step is to find the mean and variance of this random variable.
When you have the variance, $\sigma^2$, you need to find $k$ so that $k\sigma=2.5$. Then you can apply Chebychev's inequality.
Then you have $k^{-2}$ and are done; as that is the upper bound according to the inequality.
$$\Bbb P\big(\lvert X-\Bbb E(X)\rvert \geqslant 2.5\big) ~ \leqslant~ \dfrac {\sigma^2}{6.25}$$