For the first one, use some right angled trig. You'll find:

Next, we just multiply the width of the rectangle from Part (i), with the height, which we established was \(r\sin{\theta}\). You'll get:
=r^2\left(\sin{\theta}\cos{\theta}-\sqrt{3}\sin^2{\theta}\right))
The last bit requires you to differentiate this function of area and find the maximum, but the method is a little different than normal. The hardest bit is the derivative itself, which requires the product rule and the chain rule as well. You should get:
)
Now to prove that \(\theta=\frac{\pi}{12}\) corresponds to a maximum, we actually need to just substitute into this expression to prove it puts the derivative equal to zero. You'll also need to substitute values slightly smaller and larger to prove it is a max and not a min

this is the atypical bit, but we do it because finding the exact value of \(\frac{\pi}{12}\) is, if memory serves, otherwise impossible with 2U methods - So we just use their indicator to guide us

a brief rundown but hopefully it helps!