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, 


$$J(\varepsilon)=2\varepsilon\int_\mathbb{R} e^{-u^2}du\big(1+o(1)\big)=2\sqrt{\pi}\varepsilon\big(1+o(1)\big),\, \,  \mathrm{as }\,  \varepsilon\to0\, .$$