For the last part, you have to find where [tex]f'(t)[/tex] attains its maximum over [tex]0\le t\le8[/tex]. We have
[tex]f'(t)=-6\sin3t+6\cos2t[/tex]
so that
[tex]f''(t)=-18\cos3t-12\sin2t[/tex]
with critical points at [tex]t[/tex] such that
[tex]-18\cos3t-12\sin2t=0[/tex]
[tex]3\cos3t+2\sin2t=0[/tex]
[tex]3(\cos^3t-3\cos t\sin^2t)+4\sin t\cos t=0[/tex]
[tex]\cos t(3\cos^2t-9\sin^2t+4\sin t)=0[/tex]
[tex]\cos t(12\sin^2t-4\sin t-3)=0[/tex]
So either
[tex]\cos t=0\implies t=\dfrac{(2n+1)\pi}2[/tex]
or
[tex]12\sin^2t-4\sin t-3=0\implies\sin t=\dfrac{1\pm\sqrt{10}}6\implies t=\sin^{-1}\dfrac{1\pm\sqrt{10}}6+2n\pi[/tex]
where [tex]n[/tex] is any integer. We get 8 solutions over the given interval with [tex]n=0,1,2[/tex] from the first set of solutions, [tex]n=0,1[/tex] from the set of solutions where [tex]\sin t=\dfrac{1+\sqrt{10}}6[/tex], and [tex]n=1[/tex] from the set of solutions where [tex]\sin t=\dfrac{1-\sqrt{10}}6[/tex]. They are approximately
[tex]\dfrac\pi2\approx2[/tex]
[tex]\dfrac{3\pi}2\approx5[/tex]
[tex]\dfrac{5\pi}2\approx8[/tex]
[tex]\sin^{-1}\dfrac{1+\sqrt{10}}6\approx1[/tex]
[tex]2\pi+\sin^{-1}\dfrac{1+\sqrt{10}}6\approx7[/tex]
[tex]2\pi+\sin^{-1}\dfrac{1-\sqrt{10}}6\approx6[/tex]
