(a) \(\angle ABC=42+28=70^\circ\). Opposite cyclic angles give \(\angle ADC=110^\circ\). (b) Let X be the intersection of AC and BD. \(\angle AXB=180-59=121^\circ\), so in ABX, \(\angle BAX=180-121-28=31^\circ\). Then in ABC, \(\angle ACB=180-70-31=79^\circ\). ADB subtends the same chord AB, so ADB = 79°.