Two normal distributions with the same variance and two disjoint intervals

The trick is to look at derivatives. Let me assume for simplicity (but without loss of generality) that $\sigma=1$ and let $\varphi$ denote the density of a $N(0,1)$. Then, for all $t_1<t_2$, \begin{align*} \ln(\varphi(t_2 -\delta))-\ln(\varphi(t_1 -\delta))& =-\frac{1}{2}\left[(t_2-\delta)^2 - (t_1-\delta)^2\right]\\ & = -\frac{1}{2}\left[t^2-t_1^2 - 2\delta(t_2-t_1)\right]\\ & > \ln(\varphi(t_2))-\ln(\varphi(t_1)). \end{align*} Therefore, for all $t_1<t_2$, $$\varphi(t_2 - \delta)\varphi(t_1)> \varphi(t_1 - \delta)\varphi(t_2).$$ Integrating $t_2$ over $[u_2,v_2]$ ($u_2>t_1$), we obtain $$\left[F_1(v_2 - \delta)-F_1(u_2 - \delta)\right]\varphi(t_1)> \varphi(t_1 - \delta)\left[F_1(v_2)-F_1(u_2)\right].$$ The result follows by integrating $t_1$ over $[u_1,v_1]$ (with $v_1<u_2$).

Edit: you were right, my first inequality is just log-concavity of the normal distribution.