Problem 167

(Ⅰ) Since $$2\sin A\sin C\left( \frac{1}{\tan A\tan C}-1\right)=-1,$$ we have that
$$2\cos A\cos C(\tan A\tan C-1)=1$$
Hence,
$$2\cos A\cos C\left( \frac{\sin A\sin C}{\cos A\cos C}-1\right)=1,$$
which simplifies to
$$2(\sin A\sin C-\cos A\cos C)=1,$$
and thereby
$$\cos(A+C)=-\frac{1}{2}.$$
Thus, we have
$$\cos B=-\cos(A+C)=\frac{1}{2}.$$
Considering that $0 < B < \pi$, we can deduce that
$$B=\frac{\pi}{3}.$$

(Ⅱ) By the law of cosines, we have
$$\cos B= \frac{a^{2}+c^{2}-b^{2}}{2ac}=\frac{1}{2}.$$
Therefore, considering $a+c=\frac{3\sqrt{3}}{2}$ and $b=\sqrt{3}$, we get
$$\frac{(a+c)^{2}-2ac-b^{2}}{2ac}=\frac{1}{2}.$$
Substituting the given values, we have
$$\frac{27}{4}-2ac-3=ac,$$ which leads to
$$ac=\frac{5}{4}.$$
Consequently, the area of triangle $ABC$ can be calculated as
$$S_{\triangle ABC}=\frac{1}{2}ac\sin B=\frac{1}{2}\times\frac{5}{4}\times\frac{\sqrt{3}}{2}=\boxed{\frac{5\sqrt{3}}{16}}.$$