Show that $(A^T)^T = A$, where T is the conjugate transpose of the matrix.
$$\left|\psi\right\rangle = \alpha_0\left|0\right\rangle + \alpha_1\left|1\right\rangle\; \text{and}\;\left|\phi\right\rangle = \beta_0\left|0\right\rangle + \beta_1\left|1\right\rangle$$ Show that $$\left|\psi\right\rangle \otimes \left|\phi\right\rangle = \alpha_0\beta_0\left|00\right\rangle + \alpha_0\beta_1\left|01\right\rangle + \alpha_1\beta_0\left|10\right\rangle + \alpha_1\beta_1\left|11\right\rangle$$
Any help on either of these would be very much appreciated.
For the second one, you might be getting tripped up with notation. Note that $|00\rangle = |0\rangle\otimes|0\rangle$, $|01\rangle = |0\rangle\otimes|1\rangle$ and likewise for the others. Does this clear it up for you?
For the first one, we often define $A^T$ (conjugate transpose) by $\langle\psi, A\phi\rangle = \langle A^T\psi,\phi\rangle$. From this, how would we define $(A^T)^T$? (There are one or two more steps after this.)