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(B) The given limit is $l = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n} \sin \left( \frac{\pi}{4} + \frac{k \pi}{12n} \right)$.This is a R

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$\frac{6(\sqrt{2}-1)}{\pi}$

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(B) The given limit is $l = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n} \sin \left( \frac{\pi}{4} + \frac{k \pi}{12n} \right)$.This is a Riemann sum of the form $\int_{0}^{1} f(x) dx = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n} f(\frac{k}{n})$.Here,$f(x) = \sin(\frac{\pi}{4} + \frac{\pi}{12}x)$.Thus,$l = \int_{0}^{1} \sin(\frac{\pi}{4} + \frac{\pi}{12}x) dx$.Evaluating the integral:$l = \left[ -\frac{12}{\pi} \cos(\frac{\pi}{4} + \frac{\pi}{12}x) \right]_{0}^{1}$$l = -\frac{12}{\pi} \left[ \cos(\frac{\pi}{4} + \frac{\pi}{12}) - \cos(\frac{\pi}{4}) \right]$$l = -\frac{12}{\pi} \left[ \cos(\frac{4\pi}{12}) - \cos(\frac{\pi}{4}) \right]$$l = -\frac{12}{\pi} \left[ \cos(\frac{\pi}{3}) - \cos(\frac{\pi}{4}) \right]$$l = \frac{12}{\pi} \left[ \cos(\frac{\pi}{4}) - \cos(\frac{\pi}{3}) \right]$$l = \frac{12}{\pi} \left[ \frac{1}{\sqrt{2}} - \frac{1}{2} \right] = \frac{12}{\pi} \left[ \frac{\sqrt{2}-1}{2} \right] = \frac{6(\sqrt{2}-1)}{\pi}$.

$\lim _{n}$ ${\rightarrow \infty} \frac{1}{n}\left[\sin \frac{\pi}{4}+\sin \frac{\pi}{12}\left(3+\frac{1}{n}\right)+\sin \frac{\pi}{12}\left(3+\frac{2}{n}\right)+\ldots+\sin \frac{\pi}{3}\right]=$

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