Given GCF(12, 48, m) = 4 and LCM(12, 48, m) = 336, find all possible values of m.

From GCF:
For $2^x$: $\min(2, 4, x) = 2 \implies x \ge 2$.
For $3^y$: $\min(1, 1, y) = 0 \implies y = 0$.
For $7^z$: $\min(0, 0, z) = 0 \implies z \ge 0$.

From LCM:
For $2^x$: $\max(2, 4, x) = 4 \implies x \le 4$.
For $3^y$: $\max(1, 1, y) = 1 \implies y \le 1$.
For $7^z$: $\max(0, 0, z) = 1 \implies z = 1$.

Combining these:
For $x$: $x \ge 2$ and $x \le 4$. Also, $\max(2, 4, x) = 4$ means $x$ can be 2, 3, or 4.
For $y$: $y = 0$ (from GCF) and $y \le 1$ (from LCM). So, $y$ must be 0.
For $z$: $z \ge 0$ (from GCF) and $z = 1$ (from LCM). So, $z$ must be 1.

The prime factorization of $m$ is $2^x \times 3^y \times 7^z$.
Substituting the determined values for $y$ and $z$:

$$m = 2^x \times 3^0 \times 7^1 = 2^x \times 1 \times 7 = 7 \times 2^x$$

The possible values for $x$ are $2, 3, 4$.

Case 1: If $x = 2$

$$m = 7 \times 2^2 = 7 \times 4 = 28$$

Check: $\text{GCF}(12, 48, 28) = \text{GCF}(2^2 \times 3, 2^4 \times 3, 2^2 \times 7) = 2^2 = 4$.
$\text{LCM}(12, 48, 28) = \text{LCM}(2^2 \times 3, 2^4 \times 3, 2^2 \times 7) = 2^4 \times 3 \times 7 = 16 \times 21 = 336$.
This value is valid.

Case 2: If $x = 3$

$$m = 7 \times 2^3 = 7 \times 8 = 56$$

Check: $\text{GCF}(12, 48, 56) = \text{GCF}(2^2 \times 3, 2^4 \times 3, 2^3 \times 7) = 2^2 = 4$.
$\text{LCM}(12, 48, 56) = \text{LCM}(2^2 \times 3, 2^4 \times 3, 2^3 \times 7) = 2^4 \times 3 \times 7 = 16 \times 21 = 336$.
This value is valid.

Case 3: If $x = 4$

$$m = 7 \times 2^4 = 7 \times 16 = 112$$

Check: $\text{GCF}(12, 48, 112) = \text{GCF}(2^2 \times 3, 2^4 \times 3, 2^4 \times 7) = 2^2 = 4$.
$\text{LCM}(12, 48, 112) = \text{LCM}(2^2 \times 3, 2^4 \times 3, 2^4 \times 7) = 2^4 \times 3 \times 7 = 16 \times 21 = 336$.
This value is valid.

The possible values of $m$ are $\boxed{\text{28, 56, 112}}$.