The range of $$a \in R$$ for which the function $$f(x) = (4a-3)(x + \log_e 5) + 2(a-7)\cot\left(\frac{x}{2}\right)\sin^2\left(\frac{x}{2}\right)$$, $$x \neq 2n\pi$$, $$n \in N$$, has critical points, is:

We simplify the function first. Note that $$\cot\left(\frac{x}{2}\right)\sin^2\left(\frac{x}{2}\right) = \frac{\cos(x/2)}{\sin(x/2)} \cdot \sin^2\left(\frac{x}{2}\right) = \cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) = \frac{\sin x}{2}$$.

So the function becomes $$f(x) = (4a-3)(x + \log_e 5) + 2(a-7) \cdot \frac{\sin x}{2} = (4a-3)(x + \ln 5) + (a-7)\sin x$$.

Differentiating: $$f'(x) = (4a-3) + (a-7)\cos x$$. For critical points, we need $$f'(x) = 0$$, which gives $$\cos x = -\frac{4a-3}{a-7} = \frac{3-4a}{a-7}$$.

For this equation to have a solution, we need $$\left|\frac{3-4a}{a-7}\right| \leq 1$$, which means $$(3-4a)^2 \leq (a-7)^2$$. Expanding: $$9 - 24a + 16a^2 \leq a^2 - 14a + 49$$, which simplifies to $$15a^2 - 10a - 40 \leq 0$$, or $$3a^2 - 2a - 8 \leq 0$$.

Factoring: $$(3a + 4)(a - 2) \leq 0$$ (since roots are $$a = \frac{2 \pm \sqrt{4+96}}{6} = \frac{2 \pm 10}{6}$$, giving $$a = 2$$ or $$a = -\frac{4}{3}$$). This inequality holds for $$-\frac{4}{3} \leq a \leq 2$$.

Therefore, the range of $$a$$ is $$\left[-\frac{4}{3}, 2\right]$$.