(1) Denote by $Q$, $R$ the midpoints of $OB$, $OC$, respectively. It is easy to see that

$$EQ = \frac{1}{2}OB = RM, \quad MQ = \frac{1}{2}OC = RF,$$ and
$$\angle EQM = \angle EQO + \angle OQM = 2\angle EBO + \angle OQM,$$
$$\angle MRF = \angle FRO + \angle ORM = 2\angle FCO + \angle ORM.$$
Because $A$, $B$, $C$, $D$ are concyclic, and $Q$, $R$ are the midpoints of $OB$, $OC$, we have
$$\angle EBO = \angle FCO, \quad \angle OQM = \angle ORM.$$
So $\angle EQM = \angle MRF$, which implies that $\triangle EQM \cong \triangle MRF$, and $EN = FN$.
Similarly, we have $EN = FN$, so $EM \times FN = EN \times FM$ holds.

(2) Suppose that $OA = 2a$, $OB = 2b$, $OC = 2c$, $OD = 2d$ and
$$\angle OAB = \alpha, \quad \angle OBA = \beta, \quad \angle ODC = \gamma, \quad \angle OCD = \theta.$$
Then
$$\begin{aligned} \cos \angle EQM &= \cos(\angle EQO + \angle OQM) \\ &= \cos(2\beta + \angle AOB) \\ &= -\cos(\alpha - \beta). \end{aligned}$$
So
$$EM^2 = EQ^2 + QM^2 - 2EQ \times QM \times \cos\angle EQM \\ = b^2 + c^2 + 2bc \cos(\alpha - \beta).$$
Making similar equations for $EN$, $FN$, $FM$, we have
$$\begin{align*} & EN \times FM = EM \times FN \\ \Leftrightarrow & EN^2 \times FM^2 = EM^2 \times FN^2 \\ \Leftrightarrow & (a^2 + d^2 + 2ad \cos(\alpha - \beta)) \times (b^2 + c^2 + 2bc \cos(\gamma - \theta)) \\ & \qquad = (a^2 + d^2 + 2ad \cos(\gamma - \theta)) \times (b^2 + c^2 + 2bc \cos(\alpha - \beta)) \\ \Leftrightarrow & (\cos(\gamma - \theta) - \cos(\alpha - \beta))(ab - cd)(ac - bd) = 0. \end{align*}$$

Because $\alpha + \beta = \gamma + \theta$, $\cos(\gamma - \theta) - \cos(\alpha - \beta) = 0$ holds if and only if $\alpha = \gamma$, $\beta = \theta$ (i.e. $A$, $B$, $C$, $D$ are concyclic) or $\alpha = \theta$, $\beta = \gamma$ (which follows $AB \parallel CD$, a contradiction); $ab - cd = 0$ holds if and only if $AD \parallel BC$; $ac - bd = 0$ holds if and only if $A$, $B$, $C$, $D$ are concyclic.
So, when $AD \parallel BC$ holds, we also have
$$EM \times FN = EN \times FM.$$
We know that $A$, $B$, $C$, $D$ are not concyclic in this case because $PB \neq PC$, so the answer is "false".