Answer: (a) The cumulative distribution function is

\(F(x)=\mathbb{P}(X\le x)= \begin{cases} 0,&x\lt 0,\\[4pt] \dfrac{x}{8},&0\le x\lt 1,\\[6pt] \dfrac{1}{28}\left(8x-\dfrac{x^2}{2}\right)-\dfrac17,&1\le x\le 8,\\[6pt] 1,&x\gt 8. \end{cases}\)

An equivalent form for \(1\le x\le 8\) is \(F(x)=1-\dfrac{(8-x)^2}{56}\).

(b) \(a=4\).

(c) Since \(Y=\sqrt[3]{X}\), the density of \(Y\) is

\(g(y)= \begin{cases} \dfrac{3}{8}y^2,&0\le y\lt 1,\\[6pt] \dfrac{3}{28}(8y^2-y^5),&1\le y\le 2,\\[6pt] 0,&\text{otherwise}. \end{cases}\)

(a) By definition, \(F(x)=\mathbb{P}(X\le x)=\int_{-\infty}^{x} f(t)\,dt\).

We must consider the different regions of the density.

For \(x\lt 0\), there is no probability mass yet, so

\(F(x)=0\).

For \(0\le x\lt 1\),

\(F(x)=\int_0^x \frac18\,dt=\frac{x}{8}.\)

For \(1\le x\le 8\), we add the probability from \(0\) to \(1\), then integrate the second part from \(1\) to \(x\):

\(F(x)=\int_0^1 \frac18\,dt+\int_1^x \frac1{28}(8-t)\,dt.\)

The first integral is

\(\int_0^1 \frac18\,dt=\frac18.\)

The second integral is

\(\int_1^x \frac1{28}(8-t)\,dt=\frac1{28}\left[8t-\frac{t^2}{2}\right]_1^x\)

\(=\frac1{28}\left(8x-\frac{x^2}{2}-8+\frac12\right)=\frac1{28}\left(8x-\frac{x^2}{2}-\frac{15}{2}\right).\)

So

\(F(x)=\frac18+\frac1{28}\left(8x-\frac{x^2}{2}-\frac{15}{2}\right).\)

Simplifying,

\(F(x)=\frac1{28}\left(8x-\frac{x^2}{2}\right)-\frac17.\)

This can also be written as

\(F(x)=1-\frac{(8-x)^2}{56}.\)

For \(x\gt 8\), all the probability has been included, so

\(F(x)=1.\)

Therefore,

\(F(x)= \begin{cases} 0,&x\lt 0,\\[4pt] \dfrac{x}{8},&0\le x\lt 1,\\[6pt] \dfrac{1}{28}\left(8x-\dfrac{x^2}{2}\right)-\dfrac17,&1\le x\le 8,\\[6pt] 1,&x\gt 8. \end{cases}\)

(b) We need \(\mathbb{P}(X\le a)=\frac57\), so \(F(a)=\frac57\).

Since \(\frac57\) is bigger than \(F(1)=\frac18\), the value of \(a\) lies in the interval \(1\le a\le 8\). So we use the second part of \(F\):

\(\frac{1}{28}\left(8a-\frac{a^2}{2}\right)-\frac17=\frac57.\)

Add \(\frac17\) to both sides:

\(\frac{1}{28}\left(8a-\frac{a^2}{2}\right)=\frac67.\)

Multiply by \(28\):

\(8a-\frac{a^2}{2}=24.\)

Multiply by \(2\):

\(16a-a^2=48.\)

Rearrange:

\(a^2-16a+48=0.\)

Factorise:

\((a-4)(a-12)=0.\)

So \(a=4\) or \(a=12\). But \(X\) only takes values between \(0\) and \(8\), so the valid answer is

\(a=4.\)

(c) We have \(Y=\sqrt[3]{X}\), so \(X=Y^3\).

Because the cube-root function is increasing,

\(G(y)=\mathbb{P}(Y\le y)=\mathbb{P}(\sqrt[3]{X}\le y)=\mathbb{P}(X\le y^3)=F(y^3).\)

The range of \(X\) is \(0\le X\le 8\), so the range of \(Y\) is \(0\le Y\le 2\).

For \(0\le y\lt 1\), we have \(0\le y^3\lt 1\), so

\(G(y)=F(y^3)=\frac{y^3}{8}.\)

For \(1\le y\le 2\), we have \(1\le y^3\le 8\), so

\(G(y)=F(y^3)=\frac1{28}\left(8y^3-\frac{y^6}{2}\right)-\frac17.\)

Now differentiate to get the density \(g(y)=G'(y)\).

For \(0\le y\lt 1\),

\(g(y)=\frac{d}{dy}\left(\frac{y^3}{8}\right)=\frac38 y^2.\)

For \(1\le y\le 2\),

\(g(y)=\frac{d}{dy}\left(\frac1{28}\left(8y^3-\frac{y^6}{2}\right)-\frac17\right)\)

\(=\frac1{28}(24y^2-3y^5)=\frac3{28}(8y^2-y^5).\)

Outside \([0,2]\), the density is \(0\).

Hence

\(g(y)= \begin{cases} \dfrac{3}{8}y^2,&0\le y\lt 1,\\[6pt] \dfrac{3}{28}(8y^2-y^5),&1\le y\le 2,\\[6pt] 0,&\text{otherwise}. \end{cases}\)