If $$\cos \alpha + \cos \beta = \frac{3}{2}$$ and $$\sin \alpha + \sin \beta = \frac{1}{2}$$ and $$\theta$$ is the arithmetic mean of $$\alpha$$ and $$\beta$$, then $$\sin 2\theta + \cos 2\theta$$ is equal to:

We have been given the two relations

$$\cos\alpha+\cos\beta=\frac32\qquad\text{and}\qquad\sin\alpha+\sin\beta=\frac12.$$

Let us write each sum with the standard sum-to-product identities. First, we recall the identities:

$$\cos A+\cos B = 2\cos\frac{A+B}{2}\,\cos\frac{A-B}{2},$$

$$\sin A+\sin B = 2\sin\frac{A+B}{2}\,\cos\frac{A-B}{2}.$$

Putting $$A=\alpha$$ and $$B=\beta$$ we get

$$\cos\alpha+\cos\beta =2\cos\frac{\alpha+\beta}{2}\,\cos\frac{\alpha-\beta}{2} =\frac32,$$

$$\sin\alpha+\sin\beta =2\sin\frac{\alpha+\beta}{2}\,\cos\frac{\alpha-\beta}{2} =\frac12.$$

We are told that $$\theta$$ is the arithmetic mean of $$\alpha$$ and $$\beta$$, so by definition

$$\theta=\frac{\alpha+\beta}{2}.$$

For convenience we introduce another symbol for half the difference:

$$\phi=\frac{\alpha-\beta}{2}.$$

With these two abbreviations the above equations become

$$2\cos\theta\,\cos\phi=\frac32\qquad\text{and}\qquad 2\sin\theta\,\cos\phi=\frac12.$$

Simplifying each equation by dividing by 2 yields

$$\cos\theta\,\cos\phi=\frac34\qquad\text{and}\qquad \sin\theta\,\cos\phi=\frac14.$$

Because the factor $$\cos\phi$$ is common in both expressions, dividing the second equation by the first eliminates it and gives a direct relation between $$\sin\theta$$ and $$\cos\theta$$:

$$\frac{\sin\theta\,\cos\phi}{\cos\theta\,\cos\phi} =\frac{\frac14}{\frac34} \;\;\Longrightarrow\;\; \tan\theta=\frac13.$$

From $$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac13,$$ we construct a right-triangle picture: let the opposite side be 1 and the adjacent side be 3. The hypotenuse is then

$$\sqrt{1^{2}+3^{2}}=\sqrt{10}.$$

Hence

$$\sin\theta=\frac{1}{\sqrt{10}},\qquad \cos\theta=\frac{3}{\sqrt{10}}.$$

Our goal is to compute $$\sin2\theta+\cos2\theta.$$ We now recall the double-angle formulas:

$$\sin2\theta=2\sin\theta\cos\theta,\qquad \cos2\theta=\cos^{2}\theta-\sin^{2}\theta.$$

First, calculate $$\sin2\theta$$:

$$\sin2\theta =2\sin\theta\cos\theta =2\left(\frac{1}{\sqrt{10}}\right)\!\left(\frac{3}{\sqrt{10}}\right) =2\left(\frac{3}{10}\right) =\frac{6}{10} =\frac35.$$

Next, calculate $$\cos2\theta$$:

$$\cos2\theta =\cos^{2}\theta-\sin^{2}\theta =\left(\frac{3}{\sqrt{10}}\right)^{2}-\left(\frac{1}{\sqrt{10}}\right)^{2} =\frac{9}{10}-\frac{1}{10} =\frac{8}{10} =\frac45.$$

Now add the two results:

$$\sin2\theta+\cos2\theta =\frac35+\frac45 =\frac{3+4}{5} =\frac75.$$

Therefore $$\sin2\theta+\cos2\theta=\dfrac75.$$

Looking at the options, $$\dfrac75$$ corresponds to Option B.

Hence, the correct answer is Option B.