Hello Yuri,
1.) We are given the functions:
$$A(t)=-t^3+at^2+p$$
$$B(t)=bt(t-2)^2+2t^2+ct+q$$
Now, we are also given:
$$A(0)=300=-(0)^3+a(0)^2+p=p\,\therefore\,p=300$$
$$B(0)=0=b(0)(0-2)^2+2(0)^2+c(0)+q=q\,\therefore\,q=0$$
Thus, we have:
$$A(t)=-t^3+at^2+300$$
$$B(t)=bt(t-2)^2+2t^2+ct$$
Because the amount of liquid present in A plus the amount present in B has to remain constant (assuming so spills, leaks, evaporation, etc.), we know:
$$A(t)+B(t)=300$$
Which we can arrange as:
$$300-A(t)=B(t)$$
Using the definitions of the functions, we may write:
$$300-\left(-t^3+at^2+300 \right)=bt(t-2)^2+2t^2+ct$$
Distributing, expanding and collecting like terms, we have:
$$t^3-at^2=bt^3+(2-4b)t^2+(4b+c)t$$
Equating coefficients, we find:
$$b=1$$
$$4b-2=a\,\therefore\,a=2$$
$$4b+c=0\,\therefore\,c=-4$$
2.) Using the difference of cubes formula, we may write:
$$(2n+1)^3-(2n-1)^3=((2n+1)-(2n-1))\left((2n+1)^2+(2n+1)(2n-1)+(2n-1)^2 \right)=2\left(12n^2+1 \right)$$
The second factor was obtained by noting:
$$(2n+1)^2=4n^2+4n+1$$
$$(2n+1)(2n-1)=4n^2-1$$
$$(2n-1)^2=4n^2-4n+1$$
Adding, we find:
$$(2n+1)^2+(2n+1)(2n-1)+(2n-1)^2=12n^2+1$$
Since $$12n^2+1$$ is odd for any $n\in\mathbb{N}$, we find then that the difference between the cubes of two consecutive positive odd numbers can never be divisible by 4.
If we let $$m=6n^2$$ then $$12n^2+1=2m+1$$, which is obviously odd.