For the first one, let $\displaystyle t = a-s$ so the integral becomes
$\displaystyle S = -e^a \int_{a-1}^a \frac{e^{-s}}{s-a-1}\,ds$
or if your replace s with t in mine and then solve for the integral (note - you're missing a dt)
For the second, let $\displaystyle t = e^{x^2}$ (the hint is in how the limits of integration change) so the given integral becomes
$\displaystyle \alpha = \frac{1}{2} \int_e^{e^4} \frac{dt}{\sqrt{\ln t}} $
now we'll integrate by parts
$\displaystyle \frac{1}{2}\int_e^{e^4} \frac{t\,dt}{t\,\sqrt{\ln t}} $ with $\displaystyle u = t,\;dv = \frac{1}{t\,\sqrt{\ln t}} $
so
$\displaystyle \frac{1}{2} \int_e^{e^4} \frac{t\,dt}{t\,\sqrt{\ln t}} = \left. t \sqrt{\ln t}\right|_e^{e^4} - \int_e^{e^4} \sqrt{\ln t}\, dt = 2e^4 - e - \int_e^{e^4} \sqrt{\ln t}\, dt = \alpha$ and solve for your integral.
$\displaystyle t = e^{x^2} \Rightarrow \frac{dt}{dx} = 2x e^{x^2} \Rightarrow dx = \frac{dt}{2x e^{x^2}}$.
Therefore $\displaystyle \int_1^2 e^{x^2} \, dx = \int_e^{e^4} \frac{ e^{x^2} dt}{2x e^{x^2}} = \frac{1}{2} \int_e^{e^4} \frac{dt}{x}$.
But $\displaystyle t = e^{x^2} \Rightarrow \ln t = x^2 \Rightarrow \sqrt{\ln t} = x$. Substitute this into the above.