$1.$ An antiderivative of $e^{-s}$ is $-e^{-s}$, so $\int_2^\infty ke^{-s}\,ds=ke^{-2}.$ We want our integral to be $1$, that is always the case for a density. So $k=e^2$.
$2.$ The density function is zero to the left of $2$. So we want $\int_2^5 e^2 e^{-s}\,ds.$
$3.$ We want $\int_2^\infty (s)(e^2e^{-s})\,ds.$ Now do integration by parts. Note again that the density function is $0$ to the left of $2$, hence the limits. Your answer, which is negative (impossible) seems to have been obtained by integrating from $0$.
Now do integration by parts. It looks as if you got this right, apart from having a $k$, and hence an answer, twice as lage as it should be.
Remark: There is an (in this case) sloghtly easier way to calculate the expectation, that you may not have been taught. If $X$ is non-negative, with *cumulative distribution function $F_X(s)$, then $E(X)=\int_0^\infty(1-F_X(s))\,ds.$ In our case, we would really be integrating from $2$ to infinity, since the cdf is $0$ to the left of $2$.