(a) By the quotient rule, \[y\prime=\frac{(5+2\sin x)\sec^2x-2\cos x\tan x}{(5+2\sin x)^2}.\] Multiply the numerator by \(\cos^2x\): \[5+2\sin x-2\sin x\cos^2x=5+2\sin^3x.\] Therefore \[y\prime=\frac{5+2\sin^3x}{(1-\sin^2x)(5+2\sin x)^2},\quad a=5,\ b=2.\]

(b) Since \(-1\le\sin x\le1\), the numerator is at least 3, so cannot vanish. The denominator is positive wherever the curve is defined; points with \(\cos x=0\) are excluded. Hence there are no stationary points.