Answer
(a) $|A|$, $\frac{2\pi}{k}$, $b$, $y=A\cdot sin(k(t-\frac{b}{k}))$, $\frac{b}{k}$
(b) $5$, $\frac{\pi}{2}$, $\pi$, $\frac{\pi}{4}$
Work Step by Step
(a) ... the amplitude is $|A|$, the period is $\frac{2\pi}{k}$, and the phase is $b$ ...
to get $y=A\cdot sin(k(t-\frac{b}{k}))$ ... the horizontal shift is $\frac{b}{k}$
(b) the amplitude is $5$, the period is $\frac{2\pi}{4}=\frac{\pi}{2}$, the phase is $\pi$, the horizontal shift is $\frac{\pi}{4}$