The value of $$k$$ which the function $$f(x) = \begin{cases} \left(\frac{4}{5}\right)^{\frac{\tan 4x}{\tan 5x}}, & 0 < x < \frac{\pi}{2} \\ k + \frac{2}{5}, & x = \frac{\pi}{2} \end{cases}$$ is continuous at $$x = \frac{\pi}{2}$$, is

For a real-valued function to be continuous at a point, the left and right limits must exist and must equal the functional value at that point. Here the given function is

$$ f(x)= \begin{cases} \left(\dfrac{4}{5}\right)^{\dfrac{\tan 4x}{\tan 5x}}, & 0<x<\dfrac{\pi}{2} \\ k+\dfrac{2}{5}, & x=\dfrac{\pi}{2} \end{cases} $$

Because the domain on the left of $$\dfrac{\pi}{2}$$ approaches the point only from within $$\bigl(0,\dfrac{\pi}{2}\bigr)$$, we must evaluate the one-sided limit

$$ \lim_{x\to \frac{\pi}{2}^{-}}\; \left(\dfrac{4}{5}\right)^{\dfrac{\tan 4x}{\tan 5x}} . $$

Set $$x=\dfrac{\pi}{2}-h$$ with $$h>0$$ and $$h\to 0$$. This substitution keeps $$x$$ inside the open interval $$\bigl(0,\dfrac{\pi}{2}\bigr)$$ while allowing us to reach the boundary point. Rewriting the trigonometric arguments:

$$ \tan 4x=\tan\!\bigl(4\!\left(\tfrac{\pi}{2}-h\right)\bigr) =\tan\!\bigl(2\pi-4h\bigr). $$

Using the periodicity $$\tan(\theta+\pi)=\tan\theta$$ and the odd symmetry $$\tan(-\theta)=-\tan\theta$$,

$$ \tan(2\pi-4h)=\tan(-4h)=-\tan 4h. $$

Because $$h$$ is very small, the small-angle approximation $$\tan\theta\approx\theta$$ gives

$$ \tan 4h\approx 4h \quad\Longrightarrow\quad \tan 4x\approx -4h. $$

Next examine the denominator:

$$ \tan 5x=\tan\!\bigl(5\!\left(\tfrac{\pi}{2}-h\right)\bigr) =\tan\!\bigl(\tfrac{5\pi}{2}-5h\bigr). $$

Write $$\tfrac{5\pi}{2}=2\pi+\tfrac{\pi}{2}$$ and use periodicity again:

$$ \tan\!\bigl(\tfrac{5\pi}{2}-5h\bigr) =\tan\!\bigl(\tfrac{\pi}{2}-5h\bigr). $$

The identity $$\tan\!\bigl(\tfrac{\pi}{2}-\theta\bigr)=\cot\theta$$ now yields

$$ \tan 5x=\cot(5h)=\frac{1}{\tan(5h)}. $$

Apply the small-angle approximation once more: $$\tan(5h)\approx 5h$$, so

$$ \tan 5x\approx\frac{1}{5h}. $$

Assembling the ratio in the exponent,

$$ \frac{\tan 4x}{\tan 5x} \ \approx\ \frac{-4h}{\dfrac{1}{5h}} \;=\;(-4h)\,(5h) \;=\;-20h^{2}. $$

Because $$h\to 0$$, we have $$-20h^{2}\to 0^{-}$$. Therefore

$$ \lim_{x\to\frac{\pi}{2}^{-}} \left(\dfrac{4}{5}\right)^{\frac{\tan 4x}{\tan 5x}} = \left(\dfrac{4}{5}\right)^{\,\;\lim\limits_{h\to 0}(-20h^{2})} = \left(\dfrac{4}{5}\right)^{0} =1. $$

Continuity at $$x=\dfrac{\pi}{2}$$ demands that this limit equal the defined functional value there, namely $$k+\dfrac{2}{5}$$. Hence

$$ k+\frac{2}{5}=1 \quad\Longrightarrow\quad k=1-\frac{2}{5} =\frac{3}{5}. $$

Hence, the correct answer is Option D.