In part ii, I've gotten the actual function, but how do I find out the range in which this is valid? The official answer just looks at the probability density function of X , observes that it is valid from 2 to 4, and deduces that for Y it must be valid from 8 to 64. Why is this allowed, and is there any other way to deduce the limits within which the probability density function is valid?
Yes. Since $Y = X^3$ and the support of $X$ is $[2;4]$, then the support of $Y$ is $[2^3;4^3]$.
The density function for $X$ is zero when $X$ is in $(-\infty; 2)\cup(4;\infty)$. So the density function of $X^3$ will be zero at the same, which is when $X^3$ is in $(-\infty; 2^3)\cup(4^3;\infty)$. Thus...
That's pretty much it.
Because, the probability density function is the unsigned differential of the cumulative distribution function, and the CDF is obtained by $\mathsf P(8\leq Y\leq y)=\mathsf P(2\leq X\leq y^{1/3})$ .
So $f_Y(y)~{=~\begin{vmatrix}\dfrac{\mathrm d~F_Y(y)}{\mathrm d~y\qquad~}\end{vmatrix} \\[1ex]=~\begin{vmatrix}\dfrac{\mathrm d~F_X(y^{1/3})}{\mathrm d~y\qquad\quad}\end{vmatrix} \\[1ex] =~\begin{vmatrix}\dfrac{\mathrm d~~}{\mathrm d~y}\int_2^{y^{1/3}} f_X(x)\operatorname d x\end{vmatrix} \\[1ex] =~ \begin{vmatrix}\dfrac{\mathrm d~y^{1/3}}{\mathrm d~y~\quad}\end{vmatrix} f_X(y^{1/3})\\[1ex] =~\dfrac{f_X(y^{1/3})}{3y^{2/3}}}$