Converting to NAND only I've been trying to work this out for days and still can't do it.
I have to convert the top equation to NAND only. I've worked out the second line by using Demorgans theorem however doing this would never convert the equation to NAND only.

My working:
$$\overline{ A\oplus(B+C) }\\\overline{\overline A \cdot(B+C)+A\left(\overline{B+C}\right)}$$
 A: You should of course use the identity:
$$X+Y=\overline{\overline{X}\cdot\overline{Y}}$$
And an overly complex NOT gate might also be useful:
$$\overline{X}=\overline{X\cdot X}$$
A: We are basically dealing with a XNOR/coincidence logic gate, whose expression in terms of NAND gates can be found here (see picture to the right), with the observation that the “B” from the image corresponds to your $B+C$, which can be rewritten as $\overline{\bar B\cdot\bar C}$



In boolean algebra form, each NAND gate of inputs X and Y corresponds to $\overline{X\cdot Y}$, so the image above can be expressed as $\overline{\overline{\overline{\overline{A\cdot B'}\cdot A}\cdot\overline{\overline{A\cdot B'}\cdot B'}}\cdot1}$ , where B' is $B+C$. If you are not allowed to use $1$, then the expression becomes $\overline{\overline{\overline{\overline{A\cdot B'}\cdot A}\cdot\overline{\overline{A\cdot B'}\cdot B'}}\cdot\overline{\overline{\overline{A\cdot B'}\cdot A}\cdot\overline{\overline{A\cdot B'}\cdot B'}}}$ .
A: The set $\{\uparrow\}$ is functionally complete. This means that the remaining logical connectives can be expressed in terms of alternative denial. It is easy to verify that $\phi \vee \psi \equiv (\phi \uparrow \phi) \uparrow (\psi \uparrow \psi)$ and $\neg \phi \equiv \phi \uparrow \phi$, from which the remaining connectives can be constructed via the usual equivalences.
