Differentiate $I(\varepsilon):=2\int_0^\infty e^{-\varepsilon x^2}(1+x^2)^{-1/2}dx$ with respect to $\varepsilon$ under the sign of integral; change variable putting $u:=\varepsilon x^2$. We get $$I'(\varepsilon) = -\frac{1}{\varepsilon} \int_0^\infty e^{-u}\sqrt{\frac{u}{u+\varepsilon}}du= -\frac{1}{\varepsilon}\big(1+o(1)\big)\\ ,$$ by the dominated convergence theorem, and integrating $$I(\varepsilon)=-\log(\varepsilon)\big(1+o(1)\big),\\ \\ \mathrm{as }\\ \varepsilon\to0\\ .$$
For the latter integral, change variable with $x:=\varepsilon u $, so
$$J(\varepsilon):=2\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}}e^{-{\sin(x)^2}/{\varepsilon^2}}dx=2\varepsilon\int_\mathbb{R} e^{-\sin(\varepsilon u)^2/\varepsilon^2}\chi_{ [-\frac{\pi}{2\varepsilon},\\ +\frac{\pi}{2\varepsilon} ] }(u) du\\ .$$ The integrand converges pointwise to $e^{-u^2}$ as $\varepsilon\to0$, and it is dominated by $e^{-u^2/4}$ for any $u\in\mathbb{R}$ (just because $\sin(x)\ge x/2\ge 0$ for any $0\le x \le \pi/2$ ). By the dominated convergence theorem,
$$J(\varepsilon)=2\varepsilon\int_\mathbb{R} e^{-u^2}du\big(1+o(1)\big)=2\sqrt{\pi}\varepsilon\big(1+o(1)\big),\\ \\ \mathrm{as }\\ \varepsilon\to0\\ .$$