You want
$$\int_1^\infty k s^{-6}\,ds=1.$$
Integrate. You should get $\frac{k}{5}$, so $k=5$.
For $E(X)$, use the usual formula
$$E(X)=\int_{-\infty}^\infty sf_X(s)\,ds.$$
Our density function is $0$ except on $[1,\infty)$, so
$$E(X)=\int_1^\infty (s)(5s^{-6})\,ds.$$
The integrand is $5s^{-5}$. Now do the integration.
For the variance, you can use $E((X-\mu)^2)$, where $\mu=E(X)$. However, as usual it will be easier if you use the fact that the variance is $E(X^2)-(E(X))^2$. So the only thing not yet known is $E(X^2)$. But
$$E(X^2)=\int_1^\infty (s^2)(5s^{-6})\,ds.$$
Simplify the integrand, and integrate.
For the probability, if there is no typo, our interval is an interval of negative numbers. On the negatives, the density is $0$, so the integral of $f_X(s)$ over this interval is $0$.