Answer: (i) \(I_2=\frac{1}{3}(e-1)\)

(ii) \(3I_n=e-(n-2)I_{n-3}\) for \(n\ge 3\)

(iii) \(I_8=\frac{1}{3}e-\frac{4}{9}\)

We are given \(I_n=\int_0^1 x^n e^{x^3}\,dx\).

(i) For \(I_2\), use the substitution \(u=x^3\). Then \(du=3x^2\,dx\), so \(x^2\,dx=\frac13\,du\). When \(x=0\), \(u=0\); when \(x=1\), \(u=1\). Hence

\(I_2=\int_0^1 x^2 e^{x^3}\,dx=\frac13\int_0^1 e^u\,du=\frac13\bigl[e^u\bigr]_0^1=\frac13(e-1).\)

(ii) For \(n\ge 3\), write \(x^n=x^{n-2}\cdot x^2\), so

\(I_n=\int_0^1 x^{n-2}\,x^2 e^{x^3}\,dx.\)

Now let \(u=x^3\). Then \(du=3x^2\,dx\), so \(x^2\,dx=\frac13 du\), and \(x^{n-2}=(x^3)^{(n-2)/3}=u^{(n-2)/3}\) is not the most convenient form. Instead, use integration by parts in a way that lowers the power by 3:

Take \(u=x^{n-2}\) and \(dv=x^2 e^{x^3}\,dx\). Then \(du=(n-2)x^{n-3}\,dx\), and with \(w=x^3\) we have

\(v=\int x^2 e^{x^3}\,dx=\frac13 e^{x^3}.\)

So

\(I_n=\left[\frac13 x^{n-2}e^{x^3}\right]_0^1-\int_0^1 \frac13 e^{x^3}(n-2)x^{n-3}\,dx.\)

The boundary term is \(\frac13 e\) since at \(x=1\) it is \(\frac13e\), and at \(x=0\) it is \(0\) because \(n\ge 3\). Thus

\(I_n=\frac13 e-\frac{n-2}{3}I_{n-3}.\)

Multiplying by 3 gives

\(3I_n=e-(n-2)I_{n-3}.\)

(iii) We use the recurrence for \(n=8\):

\(3I_8=e-6I_5,\qquad 3I_5=e-3I_2.\)

From part (i), \(I_2=\frac13(e-1)\), so

\(3I_5=e-3\cdot\frac13(e-1)=e-(e-1)=1,\)

hence \(I_5=\frac13\). Then

\(3I_8=e-6\cdot\frac13=e-2,\)

so

\(I_8=\frac{e-2}{3}=\frac13e-\frac23.\)

But this is not yet correct, because the recurrence should be applied carefully: for \(n=5\),

\(3I_5=e-(5-2)I_2=e-3I_2.\)

Substitute \(I_2=\frac13(e-1)\):

\(3I_5=e-3\cdot\frac13(e-1)=e-(e-1)=1,\)

so \(I_5=\frac13\), which is correct. Then for \(n=8\),

\(3I_8=e-(8-2)I_5=e-6\cdot\frac13=e-2,\)

therefore

\(I_8=\frac{e-2}{3}.\)

So the exact value is \(\boxed{I_8=\frac{e-2}{3}}\).