Answer
(a) $34$ days and $308$ days
(b) $275$ days.
Work Step by Step
(a) Let $L(t)=10$, we have $12+2.83sin(\frac{2\pi}{365}(t-80))=10$ or $sin(\frac{2\pi}{365}(t-80))=-0.7067$
Since $0\leq t\leq365$, we have $-1.37\leq\frac{2\pi}{365}(t-80)\leq4.91$, thus $\frac{2\pi}{365}(t-80)=sin^{-1}(-0.7067)\approx-0.785, 3.93$ which gives $t\approx34$ days and $308$ days
(b) Test a few days between 34 and 308, we can see that $L(t)\gt10$, so the number of days of the year having more than 10 h of daylight lie between 34 and 308, which gives $308-34+1=275$ days.
