6
Period of $\sin\!\left(\tfrac{\pi t}{6}\right)$ is $\tfrac{2\pi}{\pi/6} = 12$ — exactly one period. Sine averages to $0$, so
$\bar D = 5$ m
7
$\bar T = \tfrac{1}{14}\int_0^{14}\bigl[t\sin(\tfrac{\pi t}{4})+20\bigr]dt$. CAS: $\int_0^{14} t\sin(\tfrac{\pi t}{4})\,dt = -\tfrac{16}{\pi^2}$. So $\bar T = \tfrac{1}{14}\bigl(280 - \tfrac{16}{\pi^2}\bigr) =$
$20 - \tfrac{8}{7\pi^2} \approx 19.88\,°$C
8
$\bar S = \tfrac{1}{8}[0.2t^2+t]_0^8 = \tfrac{20.8}{8} =$ $\$2.60$.
$\bar B = \tfrac{1}{8}[-0.1t^3+1.35t^2-0.83t-\cos(2t)]_0^8 = \tfrac{29.56 - \cos 16}{8} \approx$ $\$3.81$.
Blue stock averaged higher.