Prove that $\text{tr} (\phi \otimes \psi) = \text {tr} \phi \text {tr} \psi $.

One approach to compute the trace is as follows:

Suppose we have bases $\{e_1,\dots,e_n\},\{f_1,\dots,f_m\}$ and associated dual bases $\{\alpha_1,\dots,\alpha_n\},\{\beta_1,\dots,\beta_m\}$ respectively. It follows that the sets $\{e_i \otimes f_j\}$ and $\{\alpha_i \otimes \beta_j\}$ form a basis and associated dual basis for $E \otimes F$. It follows that $$ \begin{align} \operatorname{tr}(\phi \otimes \psi) &= \sum_{i=1}^n \sum_{j=1}^m (\alpha_i \otimes \beta_j)(\phi \otimes \psi)(e_i \otimes \beta_j) \\&= \sum_{i=1}^n \sum_{j=1}^m (\alpha_i \otimes \beta_j)(\phi(e_i) \otimes \psi(f_j)) \\&= \sum_{i=1}^n \sum_{j=1}^m \alpha_i(\phi(e_i)) \beta_j(\psi(f_j)) \\ & = \left(\sum_{i=1}^n \alpha_i(\phi(e_i)) \right) \left(\sum_{j=1}^m \beta_j(\psi(f_j)) \right) = \operatorname{tr}(\phi)\operatorname{tr}(\psi). \end{align} $$

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For the determinant, use the fact that for maps $\Phi,\Psi$ over $E \otimes F$, $\det(\Phi_1 \circ \Phi_2) = \det(\Phi_1) \det(\Phi_2)$. Now, define $\Phi = \phi \otimes \operatorname{id}_F$, so that $$ \Phi(x \otimes y) = \phi(x) \otimes y. $$ Similarly, define $\Psi = \operatorname{id}_E \otimes \psi$. It suffices to show that $\det(\Phi) = \det(\phi)^m$, and $\det(\Psi) = \det(\psi)^n$.

To show that $\det(\Phi) = \det(\phi)^m$, note that the spaces $V_i = \{x \otimes \beta_i : x \in E\}$ are invariant subspaces of $\Phi$. So, we can write $\Phi$ as a direct sum of maps $$ \Phi = \overbrace{\phi \oplus \cdots \oplus \phi}^m. $$ It follows that $\det\Phi = \det(\psi)^m$, which was what we wanted. The proof for $\det \Psi$ is similar. The conclusion follows.