Let $$\alpha$$ and $$\beta$$ be real numbers such that $$-\dfrac{\pi}{4} < \beta < 0 < \alpha < \dfrac{\pi}{4}$$. If $$\sin(\alpha + \beta) = \dfrac{1}{3}$$ and $$\cos(\alpha - \beta) = \dfrac{2}{3}$$, then the greatest integer less than or equal to $$\left(\dfrac{\sin\alpha}{\cos\beta} + \dfrac{\cos\beta}{\sin\alpha} + \dfrac{\cos\alpha}{\sin\beta} + \dfrac{\sin\beta}{\cos\alpha}\right)^2$$ is _______.

Let $$A=\alpha+\beta$$ and $$B=\alpha-\beta$$.
Given ranges  $$-\dfrac{\pi}{4}\lt\beta\lt 0\lt\alpha\lt\dfrac{\pi}{4}$$ imply $$0\lt A\lt\dfrac{\pi}{4}$$ and $$0\lt B\lt\dfrac{\pi}{2}$$, so the signs required below are fixed.

Data provided
$$\sin A=\dfrac13,\qquad\cos B=\dfrac23$$

1. Values for the individual functions of $$A$$ and $$B$$

$$\cos A=\sqrt{1-\sin ^2A}= \sqrt{1-\dfrac19}= \dfrac{2\sqrt2}{3}$$
$$\sin B=\sqrt{1-\cos ^2B}= \sqrt{1-\dfrac49}= \dfrac{\sqrt5}{3}$$

2. Values for $$\cos(A+B)$$ and $$\cos(A-B)$$ (addition-subtraction formulae)

$$\cos(A+B)=\cos A\cos B-\sin A\sin B =\dfrac{2\sqrt2}{3}\cdot\dfrac23-\dfrac13\cdot\dfrac{\sqrt5}{3} =\dfrac{4\sqrt2-\sqrt5}{9}\;.$$

$$\cos(A-B)=\cos A\cos B+\sin A\sin B =\dfrac{2\sqrt2}{3}\cdot\dfrac23+\dfrac13\cdot\dfrac{\sqrt5}{3} =\dfrac{4\sqrt2+\sqrt5}{9}\;.$$

3. Half-angle use to obtain $$\alpha=\dfrac{A+B}{2}$$ and $$\beta=\dfrac{A-B}{2}$$

Because $$0\lt\alpha\lt\dfrac{\pi}{4}$$ both $$\sin\alpha$$ and $$\cos\alpha$$ are positive.
Because $$-\dfrac{\pi}{4}\lt\beta\lt0$$ we have $$\cos\beta>0$$ but $$\sin\beta<0$$.

$$\sin\alpha =\sqrt{\dfrac{1-\cos2\alpha}{2}} =\sqrt{\dfrac{1-\cos(A+B)}{2}} =\sqrt{\dfrac{1-\dfrac{4\sqrt2-\sqrt5}{9}}{2}} \approx 0.5568$$

$$\cos\alpha =\sqrt{\dfrac{1+\cos2\alpha}{2}} =\sqrt{\dfrac{1+\dfrac{4\sqrt2-\sqrt5}{9}}{2}} \approx 0.8301$$

$$\sin\beta =-\sqrt{\dfrac{1-\cos2\beta}{2}} =-\sqrt{\dfrac{1-\cos(A-B)}{2}} =-\sqrt{\dfrac{1-\dfrac{4\sqrt2+\sqrt5}{9}}{2}} \approx -0.2480$$

$$\cos\beta =\sqrt{\dfrac{1+\cos2\beta}{2}} =\sqrt{\dfrac{1+\dfrac{4\sqrt2+\sqrt5}{9}}{2}} \approx 0.9687$$

4. Evaluate the four required quotients

$$\dfrac{\sin\alpha}{\cos\beta}\approx\dfrac{0.5568}{0.9687}=0.5748$$
$$\dfrac{\cos\beta}{\sin\alpha}\approx\dfrac{0.9687}{0.5568}=1.7388$$
$$\dfrac{\cos\alpha}{\sin\beta}\approx\dfrac{0.8301}{-0.2480}=-3.3489$$
$$\dfrac{\sin\beta}{\cos\alpha}\approx\dfrac{-0.2480}{0.8301}=-0.2988$$

Adding them:
$$S=\dfrac{\sin\alpha}{\cos\beta}+\dfrac{\cos\beta}{\sin\alpha} +\dfrac{\cos\alpha}{\sin\beta}+\dfrac{\sin\beta}{\cos\alpha} \approx 0.5748+1.7388-3.3489-0.2988 \approx -1.3341$$

5. Square of the sum

$$S^2\approx (-1.3341)^2\approx 1.78$$

6. The required greatest integer

The value $$S^2$$ lies between $$1$$ and $$2$$, so
$$\lfloor S^2\rfloor=1$$

Final answer: 1