= Solution
The temporal <intensity> is proportional to the squared field modulus:
$$
I(t)=I_0e^{-2at^2}.
$$
Its half-maximum points satisfy $2at^2=\ln2$, so the <intensity> <full width at half maximum> is
$$
\boxed{\tau=\sqrt{\frac{2\ln2}{a}}.}
$$
With the chosen $e^{-i\omega t}$ convention, the instantaneous <angular frequency> is minus the time derivative of the field's argument. Hence
$$
\boxed{\omega(t)=\omega_0+2bt.}
$$
A localized Gaussian requires $\operatorname{Re}\Gamma=a>0$. It is positively chirped precisely when $d\omega/dt=2\operatorname{Im}\Gamma>0$, or $b>0$; $b=0$ gives no chirp and $b<0$ gives negative chirp.
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