Let a be the area of the triangle ABC. Then the value of $$(64a)^2$$ is

Let the angles of the triangle be $$A \le B \le C$$ and the sides opposite to them be $$a,\,b,\,c$$ respectively.
Because the triangle is obtuse, $$C$$ is the obtuse angle. We are told that

$$C-A=\frac{\pi}{2} \qquad -(1)$$

All three vertices lie on a circle of radius $$R=1$$, so by the sine rule

$$a=2\sin A,\; b=2\sin B,\; c=2\sin C \qquad -(2)$$

The sides are in arithmetic progression, hence

$$2b=a+c \;\Longrightarrow\; 2\sin B=\sin A+\sin C \qquad -(3)$$

Using $$C=A+\frac{\pi}{2}$$ from $$(1)$$, we get $$\sin C=\sin\!\left(A+\frac{\pi}{2}\right)=\cos A$$. Also, since $$A+B+C=\pi$$,

$$B=\pi-A-C=\pi-A-\left(A+\frac{\pi}{2}\right)=\frac{\pi}{2}-2A \qquad -(4)$$

Now write every term in $$(3)$$ in terms of $$A$$:

$$\sin B=\sin\!\left(\frac{\pi}{2}-2A\right)=\cos 2A$$ $$\sin C=\cos A$$

Substituting in $$(3)$$,

$$2\cos 2A=\sin A+\cos A \qquad -(5)$$

Square both sides of $$(5)$$:

$$4\cos^2 2A=\sin^2 A+\cos^2 A+2\sin A\cos A=1+\sin 2A \qquad -(6)$$

Rewrite $$\cos^2 2A=\dfrac{1+\cos 4A}{2}$$ and put this in $$(6)$$:

$$2\bigl(1+\cos 4A\bigr)=1+\sin 2A\;\;\Longrightarrow\;\;1+2\cos 4A=\sin 2A \qquad -(7)$$

Set $$t=2A$$ (note $$0\lt t\lt \dfrac{\pi}{2}$$ because $$A\lt \dfrac{\pi}{4}$$). Equation $$(7)$$ becomes

$$1+2\cos 2t=\sin t \qquad -(8)$$

Using $$\cos 2t=1-2\sin^2 t$$, convert $$(8)$$ entirely to $$\sin t$$:

$$1+2\bigl(1-2\sin^2 t\bigr)=\sin t \;\;\Longrightarrow\;\;3-4\sin^2 t-\sin t=0$$

This is a quadratic in $$\sin t$$:

$$4\sin^2 t+\sin t-3=0 \qquad -(9)$$

Solving $$(9)$$,

$$\sin t=\frac{-1\pm\sqrt{1+48}}{8}=\frac{-1\pm7}{8}$$

The negative root is extraneous, hence

$$\sin t=\frac{3}{4} \;\;\Longrightarrow\;\; \sin 2A=\frac{3}{4} \qquad -(10)$$

Because $$0\lt 2A\lt \dfrac{\pi}{2}$$, $$\cos 2A=\sqrt{1-\left(\frac{3}{4}\right)^2}=\frac{\sqrt7}{4} \qquad -(11)$$

Half-angle formulae give

$$\sin A=\sqrt{\frac{1-\cos 2A}{2}}=\sqrt{\frac{1-\dfrac{\sqrt7}{4}}{2}} \;=\;\sqrt{\frac{4-\sqrt7}{8}}$$ $$\cos A=\sqrt{\frac{1+\cos 2A}{2}}=\sqrt{\frac{1+\dfrac{\sqrt7}{4}}{2}} \;=\;\sqrt{\frac{4+\sqrt7}{8}} \qquad -(12)$$

From $$(11)$$, $$\sin B=\cos 2A=\dfrac{\sqrt7}{4}$$ and from $$(12)$$, $$\sin C=\cos A$$.

Area of a triangle with circum-radius $$R$$ is $$\Delta=\dfrac{abc}{4R}$$. Using $$R=1$$ and $$(2)$$,

$$\Delta=2\sin A\sin B\sin C=2(\sin A)(\cos 2A)(\cos A) \qquad -(13)$$

Compute the product $$\sin A\cos A$$ from $$(12)$$:

$$\sin A\cos A=\sqrt{\frac{4-\sqrt7}{8}}\, \sqrt{\frac{4+\sqrt7}{8}} =\frac{\sqrt{(4-\sqrt7)(4+\sqrt7)}}{8} =\frac{\sqrt{16-7}}{8}=\frac{3}{8} \qquad -(14)$$

Put $$(11)$$ and $$(14)$$ in $$(13)$$:

$$\Delta=2\left(\frac{3}{8}\right)\left(\frac{\sqrt7}{4}\right) =\frac{3\sqrt7}{16} \qquad -(15)$$

Finally,

$$64\Delta = 64\left(\frac{3\sqrt7}{16}\right)=12\sqrt7$$ $$\bigl(64\Delta\bigr)^2=(12\sqrt7)^2=144\times7=1008$$

Therefore, the required value is 1008.