Calculate the ratio of the areas of the circular regions below, knowing that the measurement of the ATB arc plus the measurement of the APB arc equals 450 degrees. (answer:$\frac{1}{2}$)
I try
$\alpha + \theta = 450^o \\ 360^o -\alpha + 360^o -\theta + \angle A + \angle B = 360^o\\ 720^o -(\alpha+\beta)+\angle A + \angle B = 360^o \implies 720^o -450^o +\angle A + \angle B = 360^o \\ \therefore \angle A + \angle B = 90^o $
Unfortunately, the problem does not explicitly state an important detail (although its diagram does indicate it, as I explain below), and it's also somewhat poorly worded. Nonetheless, using the values in your diagram, note the original diagram shows the circle containing $P$, with center $O$ and, say, radius $r_1$, goes through the center of the circle containing $T$ (this comes from the dotted line, indicating the radius, going from the first to the second circle), with center $C$ and, say, radius $r_2$.
Joining $O$ to $C$, then since $r_1 = \lvert OA\rvert = \lvert OB\rvert$ and $r_2 = \lvert CA\rvert = \lvert CB\rvert$, we get by SSS that
$$\triangle OAC \cong \triangle OBC$$
Using this and your result of $\measuredangle A + \measuredangle B = 90^{\circ}$, then
$$\measuredangle A = \measuredangle B = 45^{\circ}$$
Since $\lvert OA\rvert = \lvert OC\rvert$, this means $\triangle OAC$ is isosceles. Thus, we have
$$\measuredangle AOC = \measuredangle BOC = 90^{\circ}$$
This matches what's suggested in the original diagram, i.e., that $AOB$ is a straight line. In addition, we also get from $\triangle OAC$ being a right-angled triangle that
$$\lvert CA\rvert = \sqrt{2}\,\lvert AO\rvert \;\;\to\;\; r_2 = \sqrt{2}\,r_1$$
Thus, the ratio of the circle areas (that the question calls "circular regions", and which I initially interpreted as meaning some portion(s) of the circles, e.g., the different shaded regions), of the smaller circle to the larger one (the question also doesn't specify this order, but it's implicit based on the stated answer), is
$$\frac{\pi\,r_1^2}{\pi\,r_2^2} = \frac{r_1^2}{(\sqrt{2}\,r_1)^2} = \frac{1}{2}$$