(a) To find the y-coordinates of points \(A\) and \(B\), substitute \(x = 0\) into the circle's equation:

\((0-3)^2 + (y-5)^2 = 40\)

\(9 + (y-5)^2 = 40\)

\((y-5)^2 = 31\)

\(y - 5 = \pm \sqrt{31}\)

\(y = 5 \pm \sqrt{31}\)

(b) The midpoint of \(A\) and \(B\) is the center of the new circle. The y-coordinates of \(A\) and \(B\) are \(5 + \sqrt{31}\) and \(5 - \sqrt{31}\), so the midpoint is:

\(\left(0, \frac{(5 + \sqrt{31}) + (5 - \sqrt{31})}{2}\right) = (0, 5)\)

The diameter of the circle is the distance between \(A\) and \(B\), which is:

\(2\sqrt{31}\)

The radius is half of the diameter:

\(\sqrt{31}\)

The equation of the circle with center \((0, 5)\) and radius \(\sqrt{31}\) is:

\(x^2 + (y-5)^2 = 31\)