a)
Use AM-GM and $xyz=8$ to get
$$x^2 + y^2 + z^2 + xy + xy + xz + xz + yz + yz \geq 9\sqrt{x^6 y^6 z^6} = 36.$$

We have equality for $x = y = z = 2$.

b)
Using $xyz = 8$, we can transform the given expression:
$$x^2 + y^2 + z^2 + mxy + nxz + pyz = x^2 + \frac{8p}{x} + y^2 + \frac{8n}{y} + z^2 + \frac{8m}{z}.$$
Since all numbers are positive reals, we can apply AM-GM inequality to get:
$$x^2 + \frac{8p}{x} = x^2 + \frac{4p}{x} + \frac{4p}{x} \geq 6\sqrt{p^2}.$$
When we apply the same procedure for $x, y, z$ and sum the inequalities, we get:
$$x^2 + y^2 + z^2 + mxy + nxz + pyz = x^2 + \frac{8p}{x} + y^2 + \frac{8n}{y} + z^2 + \frac{8m}{z} \geq 6\sqrt[3]{2}(\sqrt[3]{m^2} + \sqrt[3]{n^2} + \sqrt[3]{p^2}).$$
In order to get equality, we must have equality in all above inequalities and that happens for
$$x = \sqrt[3]{4p}, \quad y = \sqrt[3]{4n}, \quad z = \sqrt[3]{4m}.$$

We only present solution for b) part here, marking scheme for a) part is the same as in first solution. We use weighted AM-GM:
$$\begin{aligned} x^2 + y^2 + z^2 + mxy + nxz + pyz &= \sqrt[3]{p^2} \frac{x^2}{\sqrt[3]{p^2}} + \sqrt[3]{n^2} \frac{y^2}{\sqrt[3]{n^2}} + \sqrt[3]{m^2} \frac{z^2}{\sqrt[3]{m^2}} + 2\sqrt[3]{m^2} \frac{mxy}{2\sqrt[3]{m^2}} + 2\sqrt[3]{n^2} \frac{nxz}{2\sqrt[3]{n^2}} + 2\sqrt[3]{p^2} \frac{pyz}{2\sqrt[3]{p^2}} \ &\geq 3(\sqrt[3]{m^2} + \sqrt[3]{n^2} + \sqrt[3]{p^2}) \cdot 3(\sqrt[3]{m^2} + \sqrt[3]{n^2} + \sqrt[3]{p^2}) \sqrt[3]{\frac{x^2}{\sqrt[3]{p^2}}} \sqrt[3]{\frac{y^2}{\sqrt[3]{n^2}}} \sqrt[3]{\frac{z^2}{\sqrt[3]{m^2}}} \ &= 3(\sqrt[3]{m^2} + \sqrt[3]{n^2} + \sqrt[3]{p^2}) \sqrt[3]{\frac{(3\sqrt{mxy})^2}{2}} \sqrt[3]{\frac{(3\sqrt{nxz})^2}{2}} \sqrt[3]{\frac{(3\sqrt{pyz})^2}{2}} \ &= 3(\sqrt[3]{m^2} + \sqrt[3]{n^2} + \sqrt[3]{p^2}) (\sqrt[3]{\frac{(xyz)^2}{2}})^2 \ &= 3(\sqrt[3]{m^2} + \sqrt[3]{n^2} + \sqrt[3]{p^2}) \sqrt[3]{\frac{(xyz)^2}{2}} \ &= 3(\sqrt[3]{m^2} + \sqrt[3]{n^2} + \sqrt[3]{p^2}) \sqrt[3]{4^2} \ &= 6\sqrt[3]{2}(\sqrt[3]{m^2} + \sqrt[3]{n^2} + \sqrt[3]{p^2}) \end{aligned}$$
We have shown that the minimum value the expression can take is $6\sqrt[3]{2}(\sqrt[3]{m^2} + \sqrt[3]{n^2} + \sqrt[3]{p^2})$. Equality can only be achieved when $x = \sqrt[3]{4p}, y = \sqrt[3]{4n}, z = \sqrt[3]{4m}$.