first off, let's recall that a cube is just 6 squares stacked up to each other, like in the picture below. Since we know its volume, we can find how long each side is.
part A)
[tex] \bf \textit{volume of a cube}\\\\V=x^3~~\begin{cases}x=side\\[-0.5em]\hrulefill\\V=64\end{cases}\implies 64=x^3\implies \sqrt[3]{64}=x\implies 4=x [/tex]
part B)
they have a painting with an area of 12 ft², will the painting fit flat against a side? Well, it can only fit flat if the sides of the painting are the same length or smaller than the sides of the crate, we know the crate is a 4x4x4, so are the painting's sides 4 or less?
[tex] \bf \textit{area of a square}\\\\A=s^2~~\begin{cases}s=side\\[-0.5em]\hrulefill\\A=12\end{cases}\implies 12=s^2\implies \sqrt{12}=s\implies \stackrel{yes}{3.464\approx s} [/tex]
