If $\sin \alpha+\sin \beta=\frac{\sqrt{2}}{\sqrt{3}}$ and $\cos \alpha+\cos \beta=\frac{1}{\sqrt{3}}$, then the value of $\left(20 \cos \left(\frac{\alpha-\beta}{2}\right)\right)^{2}$ is _________.

We know:

$\sin \alpha + \sin \beta = \frac{\sqrt{2}}{\sqrt{3}}$

$\cos \alpha + \cos \beta = \frac{1}{\sqrt{3}}$

Using trigonometric identities:

$\sin \alpha + \sin \beta = 2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$ ...eq.1

$\cos \alpha + \cos \beta = 2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$ .....eq.2

This gives us:

$2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right) = \frac{\sqrt{2}}{\sqrt{3}}$

$2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right) = \frac{1}{\sqrt{3}}$

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Dividing these equations:

$\tan\left(\frac{\alpha+\beta}{2}\right) = \frac{\sin\left(\frac{\alpha+\beta}{2}\right)}{\cos\left(\frac{\alpha+\beta}{2}\right)} = \frac{\frac{\sqrt{2}}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = \sqrt{2}$

From the second equation:

$\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right) = \frac{1}{2\sqrt{3}}$

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Since $\tan\left(\frac{\alpha+\beta}{2}\right) = \sqrt{2}$, we get:

$\cos\left(\frac{\alpha+\beta}{2}\right) = \frac{1}{\sqrt{1+\tan^2\left(\frac{\alpha+\beta}{2}\right)}} = \frac{1}{\sqrt{3}}$

Substituting:

$\frac{1}{\sqrt{3}} \cdot \cos\left(\frac{\alpha-\beta}{2}\right) = \frac{1}{2\sqrt{3}}$

$\cos\left(\frac{\alpha-\beta}{2}\right) = \frac{1}{2}$

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Therefore:

$\left(20 \cos\left(\frac{\alpha-\beta}{2}\right)\right)^2 = \left(20 \cdot \frac{1}{2}\right)^2 = 10^2 = 100$