In the Foreman-Woodin model (mathscinet MR1087344), for all $\kappa$, $\beth_n(\kappa)$ is at least the $n^{th}$ weakly inaccessible above $\kappa$. Furthermore, this model satisfies the following:

(1) $\forall \kappa \forall n (2^\kappa > \kappa^{+n})$

(2) $\forall \kappa \forall n (2^\kappa = 2^{\kappa^{+n}})$

From these statements, it follows that $\text{FCH}^m_n$ holds for all $m,n<\omega$. To show this, first note that $\text{FCH}^m_0$ holds for all $m$ by (1). Now, (2) says that $\beth_1(\kappa^{+m}) = \beth_1(\kappa)$ for all $m$. Thus $\beth_n(\kappa^{+m}) = \beth_n(\kappa)$ for all $n \geq 1$. Thus $\beth_n(\kappa^{+m}) < \beth_{n+1}(\kappa)$ by Cantor's Theorem.

In the Foreman-Woodin model (mathscinet MR1087344), for all $\kappa$, $\beth_n(\kappa)$ is at least the $n^{th}$ weakly inaccessible above $\kappa$. Furthermore, this model satisfies the following:

(1) $\forall \kappa \forall n (2^\kappa > \kappa^{+n})$

(2) $\forall \kappa \forall n (2^\kappa = 2^{\kappa^{+n}})$

From these statements, it follows that $\text{FCH}^m_n$ holds for all $m,n<\omega$. To show this, first note that $\text{FCH}^m_0$ holds for all $m$ by (1). Now, (2) says that $\beth_1(\kappa^{+m}) = \beth_1(\kappa)$ for all $m$. Thus $\beth_n(\kappa^{+m}) = \beth_n(\kappa)$ for all $n \geq 1$. Thus $\beth_n(\kappa^{+m}) < \beth_{n+1}(\kappa)$ by Cantor's Theorem.

In the Foreman-Woodin model (mathscinet MR1087344), for all $\kappa$, $\beth_n(\kappa)$ is at least the $n^{th}$ weakly inaccessible above $\kappa$. Thus this model satisfies the following:

(1) $\forall \kappa \forall n (2^\kappa > \kappa^{+n})$

(2) $\forall \kappa \forall n (2^\kappa = 2^{\kappa^{+n}})$

From these statements, it follows that $\text{FCH}^m_n$ holds for all $m,n<\omega$. To show this, first note that $\text{FCH}^m_0$ holds for all $m$ by (1). Now, (2) says that $\beth_1(\kappa^{+m}) = \beth_1(\kappa)$ for all $m$. Thus $\beth_n(\kappa^{+m}) = \beth_n(\kappa)$ for all $n \geq 1$. Thus $\beth_n(\kappa^{+m}) < \beth_{n+1}(\kappa)$ by Cantor's Theorem.

In the Foreman-Woodin model (mathscinet MR1087344), for all $\kappa$, $\beth_n(\kappa)$ is at least the $n^{th}$ weakly inaccessible above $\kappa$. Furthermore, this model satisfies the following:

(1) $\forall \kappa \forall n (2^\kappa > \kappa^{+n})$

(2) $\forall \kappa \forall n (2^\kappa = 2^{\kappa^{+n}})$

From these statements, it follows that $\text{FCH}^m_n$ holds for all $m,n<\omega$. To show this, first note that $\text{FCH}^m_0$ holds for all $m$ by (1). Now, (2) says that $\beth_1(\kappa^{+m}) = \beth_1(\kappa)$ for all $m$. Thus $\beth_n(\kappa^{+m}) = \beth_n(\kappa)$ for all $n \geq 1$. Thus $\beth_n(\kappa^{+m}) < \beth_{n+1}(\kappa)$ by Cantor's Theorem.