(i) Differentiate [tex]s[/tex] with respect to [tex]t[/tex] to recover the particle's velocity function.
[tex]s(t) = 1 + 3t - \cos(5t) \implies s'(t) = 3 + 5 \sin(t)[/tex]
The is at rest any at time [tex]t[/tex] such that [tex]s'(t)=0[/tex]. Solve for [tex]t[/tex].
[tex]3 + 5 \sin(t) = 0 \implies \sin(t) = -\dfrac35 \\\\ \implies t = \sin^{-1}\left(-\dfrac35\right) + 2n\pi \text{ or } \pi - \sin^{-1}\left(\dfrac35\right) + 2n\pi[/tex]
where [tex]n\in\Bbb Z[/tex]. The first two times this happens occurs when
[tex]t_1 = \pi - \sin^{-1}\left(-\dfrac35\right) \approx 3.78509[/tex]
and
[tex]t_2 = \sin^{-1}\left(-\dfrac35\right) + 2\pi \approx 5.63968[/tex]
Then the distance between these positions is
[tex]|s(t_2) - s(t_1)| \approx |18.9162 - 11.3582| \approx 7.558[/tex]
(Don't forget units.)
(ii) Not sure if you meant to specify a numerical value of [tex]t[/tex], like [tex]t=\pi[/tex], or an arbitrary time [tex]t=T[/tex]. Either way, just differentiate the velocity function to get the acceleration, and evaluate at this time.
[tex]s'(t) = 3 + 5 \sin(t) \implies s''(t) = 5\cos(t)[/tex]
If [tex]t=\pi[/tex], then [tex]5\cos(\pi)=-5[/tex], for instance.
