4. We know that $-1\le \cos\le 1$. So the maximum population is attained when $\cos(0.06\pi t)=1$, for example at $t=0$, and it is $P=250$. And the minimum is attained when $\cos(0.06\pi t)=-1$, for example at $t=100/6$, and it is $P=150$.

5. Suppose the leftmost vertex of the triangle is $(0,0)$ and the lower-left vertex of the rectangle is $(a,0)$. Then the other vertices of rectangle are $(a,2),(a+6,0),(a+6,2)$. One side of triangle is on the line $y=0$ and the other one is on the line $y=2x/a$ (passing through $(0,0),(a,2)$). Suppose $(l,0)$ is also a vertex of the triangle (note that $l> a+6>6$). Then the third side of the triangle lies on the line passing through $(l,0),(a+6,2)$, which is $y={2\over a+6-l}(x-l)$. So the third vertex of the triangle is $${2x\over a}={2\over a+6-l}(x-l)\implies x={al\over l-6}\implies y={2x\over a}={2l\over l-6}$$

So the area of triangle is $$ A={1\over 2}l{2l\over l-6}={l^2\over l-6}={l^2-36+36\over l-6}=l+6+{36\over l-6} \\\;\\ \implies {dA\over dl}=1-{36\over (l-6)^2}=0\implies (l-6)^2=36\implies l=6\pm 6=12 \textrm{ or } 0$$But $l>6$ so the minimum happens at $l=12$. Note that ${dA\over dl}<0 $ for $l\in (6,12)$ and ${dA\over dl}>0 $ for $l\in (12,\infty)$, so $l=12$ is indeed the minimum. And the minimum area is $$A={12^2\over12-6}=24 $$