Solution:

Here $AB = 4 + 16 + 8 = 28$ and $AC = 13 + 52 + 26 = 91,$ so $D$ and $F$ lie $\frac{1}{7}$ of the way from $A$ along their sides while $E$ and $G$ lie $\frac{5}{7}$ of the way. Triangles sharing angle $A$ have areas proportional to the products of the adjacent sides, so $[ADF] = \frac{1}{49}[ABC]$ and $[AEG] = \frac{25}{49}[ABC].$ Therefore $$[DEGF] = [AEG] - [ADF] = \frac{24}{49}[ABC] = 288,$$ which gives $[ABC] = 588.$

Now set $\mathbf{b} = \overrightarrow{AB}$ and $\mathbf{c} = \overrightarrow{AC},$ so that $D = \frac{1}{7}\mathbf{b},$ $E = \frac{5}{7}\mathbf{b},$ $F = \frac{1}{7}\mathbf{c},$ $G = \frac{5}{7}\mathbf{c},$ and the reflections are $M = 2F - D = \frac{1}{7}(2\mathbf{c} - \mathbf{b})$ and $N = 2E - G = \frac{1}{7}(10\mathbf{b} - 5\mathbf{c}).$ The shoelace formula for $AFNBCEM$ sums cross products of consecutive vertices: the two terms at $A$ vanish, and $$F \times N = -\tfrac{10}{49}\,\mathbf{b} \times \mathbf{c}, \quad N \times B = \tfrac{5}{7}\,\mathbf{b} \times \mathbf{c}, \quad B \times C = \mathbf{b} \times \mathbf{c}, \quad C \times E = -\tfrac{5}{7}\,\mathbf{b} \times \mathbf{c}, \quad E \times M = \tfrac{10}{49}\,\mathbf{b} \times \mathbf{c}.$$

Everything cancels except the single term $\mathbf{b} \times \mathbf{c},$ so the heptagon's area is $\frac{1}{2}\left|\mathbf{b} \times \mathbf{c}\right| = [ABC] = 588.$