The value of $$\cos^3\left(\frac{\pi}{8}\right) \cdot \cos\left(\frac{3\pi}{8}\right) + \sin^3\left(\frac{\pi}{8}\right) \cdot \sin\left(\frac{3\pi}{8}\right)$$ is:

We have to evaluate

$$$\cos^{3}\!\left(\frac{\pi}{8}\right)\,\cos\!\left(\frac{3\pi}{8}\right)\;+\;\sin^{3}\!\left(\frac{\pi}{8}\right)\,\sin\!\left(\frac{3\pi}{8}\right).$$$

First, we recall the standard power-reduction formulas:

$$\cos^{3}\theta=\frac{1}{4}\bigl(3\cos\theta+\cos3\theta\bigr),$$

$$\sin^{3}\theta=\frac{1}{4}\bigl(3\sin\theta-\sin3\theta\bigr).$$

We apply these with $$\theta=\frac{\pi}{8}.$$ Substituting, we obtain

$$\cos^{3}\!\left(\frac{\pi}{8}\right)=\frac{1}{4}\Bigl(3\cos\!\left(\frac{\pi}{8}\right)+\cos\!\left(\frac{3\pi}{8}\right)\Bigr),$$

$$\sin^{3}\!\left(\frac{\pi}{8}\right)=\frac{1}{4}\Bigl(3\sin\!\left(\frac{\pi}{8}\right)-\sin\!\left(\frac{3\pi}{8}\right)\Bigr).$$

Now we multiply each of these by the corresponding cosine or sine factor present in the original expression:

$$$\cos^{3}\!\left(\frac{\pi}{8}\right)\cos\!\left(\frac{3\pi}{8}\right) =\frac{1}{4}\Bigl(3\cos\!\left(\frac{\pi}{8}\right)+\cos\!\left(\frac{3\pi}{8}\right)\Bigr) \cos\!\left(\frac{3\pi}{8}\right),$$$

$$$\sin^{3}\!\left(\frac{\pi}{8}\right)\sin\!\left(\frac{3\pi}{8}\right) =\frac{1}{4}\Bigl(3\sin\!\left(\frac{\pi}{8}\right)-\sin\!\left(\frac{3\pi}{8}\right)\Bigr) \sin\!\left(\frac{3\pi}{8}\right).$$$

Adding the two products gives

$$$\begin{aligned} E &= \frac{1}{4}\Bigl(3\cos\!\left(\frac{\pi}{8}\right)+\cos\!\left(\frac{3\pi}{8}\right)\Bigr) \cos\!\left(\frac{3\pi}{8}\right)\\[4pt] &\quad+\frac{1}{4}\Bigl(3\sin\!\left(\frac{\pi}{8}\right)-\sin\!\left(\frac{3\pi}{8}\right)\Bigr) \sin\!\left(\frac{3\pi}{8}\right)\\[6pt] &=\frac{1}{4}\Bigl[ 3\cos\!\left(\frac{\pi}{8}\right)\cos\!\left(\frac{3\pi}{8}\right) +\cos^{2}\!\left(\frac{3\pi}{8}\right) +3\sin\!\left(\frac{\pi}{8}\right)\sin\!\left(\frac{3\pi}{8}\right) -\sin^{2}\!\left(\frac{3\pi}{8}\right)\Bigr]. \end{aligned}$$$

We notice in the bracket two distinct parts. For the terms with the coefficient 3 we use the angle-difference identity

$$\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos(\alpha-\beta).$$

Taking $$\alpha=\frac{3\pi}{8},\;\beta=\frac{\pi}{8},$$ we get

$$$\cos\!\left(\frac{3\pi}{8}\right)\cos\!\left(\frac{\pi}{8}\right) +\sin\!\left(\frac{3\pi}{8}\right)\sin\!\left(\frac{\pi}{8}\right) =\cos\!\left(\frac{3\pi}{8}-\frac{\pi}{8}\right) =\cos\!\left(\frac{2\pi}{8}\right) =\cos\!\left(\frac{\pi}{4}\right).$$$

So the portion with the coefficient 3 simplifies to

$$3\cos\!\left(\frac{\pi}{4}\right).$$

Next, the remaining pair $$\cos^{2}\!\left(\frac{3\pi}{8}\right)-\sin^{2}\!\left(\frac{3\pi}{8}\right)$$ can be recognized through the double-angle identity

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

Putting $$\theta=\frac{3\pi}{8},$$ we have

$$$\cos^{2}\!\left(\frac{3\pi}{8}\right)-\sin^{2}\!\left(\frac{3\pi}{8}\right) =\cos\!\left(2\cdot\frac{3\pi}{8}\right) =\cos\!\left(\frac{6\pi}{8}\right) =\cos\!\left(\frac{3\pi}{4}\right).$$$

Collecting these results, the bracketed expression becomes

$$3\cos\!\left(\frac{\pi}{4}\right)+\cos\!\left(\frac{3\pi}{4}\right).$$

Therefore,

$$E=\frac{1}{4}\Bigl[3\cos\!\left(\frac{\pi}{4}\right)+\cos\!\left(\frac{3\pi}{4}\right)\Bigr].$$

We now substitute the numeric values:

$$\cos\!\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt2},\qquad \cos\!\left(\frac{3\pi}{4}\right)=-\frac{1}{\sqrt2}.$$

Thus,

$$$E=\frac{1}{4}\Bigl[3\!\left(\frac{1}{\sqrt2}\right)+\!\left(-\frac{1}{\sqrt2}\right)\Bigr] =\frac{1}{4}\!\left(\frac{3-1}{\sqrt2}\right) =\frac{1}{4}\!\left(\frac{2}{\sqrt2}\right) =\frac{2}{4\sqrt2} =\frac{1}{2\sqrt2}.$$$

Hence, the correct answer is Option B.