$\psi_\alpha$, $\alpha=1,2,\ldots d$, is a column vector of a $d\times d$ unitary matrix $U$; averaging over the Haar measure gives $$\int d\psi\, \psi_{\alpha} \psi^\ast_\beta \psi_{\alpha'}\psi^\ast_{\beta'}=\frac{1}{d+d^2}\left(\delta_{\alpha\beta}\delta_{\alpha'\beta'}+\delta_{\alpha\beta'}\delta_{\alpha'\beta}\right)$$
so your integral equals
$$\int \mathrm{Tr}( \psi \psi^* \, \, w x^* \,\, \psi \psi^* \,\, y z^*) d\psi= \langle z|\psi\rangle\langle \psi|w\rangle\langle x|\psi\rangle\langle \psi|y\rangle$$ $$\qquad\qquad = \frac{1}{d+d^2}\bigl(\langle z|w\rangle\langle x|y\rangle+ \langle z|y\rangle\langle x|w\rangle\bigr)$$
the answer to your second question is zero, the average of $\langle\psi|x\rangle\langle y|\psi'\rangle$ with $\psi,\psi'$ two different column vectors of $U$ is zero (obviously, because $\psi,\psi'$ and $-\psi,\psi'$ are equally probable)