For $$0 < \theta < \dfrac{\pi}{4}$$, let
$$a = ((\sin \theta)^{\sin \theta}) (\log_{2} \cos \theta)$$, $$b = ((\cos \theta)^{\sin \theta}) (\log_{2} \sin \theta)$$
$$c = ((\sin \theta)^{\cos \theta}) (\log_{2} \cos \theta)$$ and $$d = ((\sin \theta)^{\sin \theta}) (\log_{2} \sin \theta)$$
Then, the median value in the sequence a, b, c, d is

We know when $$0 < \theta < \frac{\pi}{4}$$, $$\cos\theta\ >\sin\theta\ $$

Also, all the terms $$a,b,c,d$$ are multiplication of an exponential term (which is always positive) and a logarithmic term either $$\log_2\cos\theta\ $$ or $$\log_2\sin\theta\ $$

Now, both $$\sin\theta\ $$and $$\cos\theta\ $$ are fractional values in $$0 < \theta < \dfrac{\pi}{4}$$ range

So, both $$\log_2\cos\theta\ $$ and $$\log_2\sin\theta\ $$ will definitely be negative

So, $$a,b,c,d$$ all are negative terms

Now, let's compare $$a,b,c,d$$

i.) $$a$$ and $$d$$:

$$\dfrac{a}{d}=\dfrac{((\sin\theta)^{\sin\theta})(\log_2\cos\theta)}{((\sin\theta)^{\sin\theta})(\log_2\sin\theta)}$$

or, $$\dfrac{a}{d}=\dfrac{(\log_2\cos\theta)}{(\log_2\sin\theta)}$$

Since, $$\cos\theta\ >\sin\theta\ $$

or, $$\log_2\cos\theta\ >\log_2\sin\theta\ $$  (as $$\log_ax$$ is an increasing function for $$a>1$$)

or, $$\dfrac{\log_2\cos\theta}{\log_2\sin\theta}<1$$ (as both $$\log_2\cos\theta\ $$ and $$\log_2\sin\theta\ $$ are negative)

or, $$\dfrac{a}{d}<1$$
or, $$a>d$$ (as  $$a$$ and $$d$$ both are negative) ------>(1)

ii.) $$b$$ and $$d$$:

$$\dfrac{b}{d}=\dfrac{((\cos\theta)^{\sin\theta})(\log_2\sin\theta)}{((\sin\theta)^{\sin\theta})(\log_2\sin\theta)}$$

or, $$\dfrac{b}{d}=\dfrac{((\cos\theta)^{\sin\theta})}{((\sin\theta)^{\sin\theta})}=\left(\cot\ \theta\ \right)^{\sin\ \theta\ }$$

Now for $$0 < \theta < \frac{\pi}{4}$$, $$\cot\theta\ $$ greater than 1 always

So, $$\dfrac{b}{d}>1$$ 

or, $$d>b$$ (as $$b$$ and $$d$$ both are negative) ------->(2)

iii.) $$a$$ and $$c$$:

$$\dfrac{a}{c}=\dfrac{((\sin\theta)^{\sin\theta})(\log_2\cos\theta)}{((\sin\theta)^{\cos\theta})(\log_2\cos\theta)}$$

or, $$\dfrac{a}{c}=\left(\sin\theta\ \right)^{\sin\ \theta\ -\cos\ \theta\ }$$

Now, $$\sin\theta\ -\cos\theta\ $$ is a negative term and $$\sin\theta$$ has a fractional value

So, $$\left(\sin\theta\right)^{\sin\theta\ -\cos\theta\ }$$ will definitely be greater than 1 (as fractional term having negative power is always greater than 1)

So, $$\dfrac{a}{c}>1$$

or, $$c>a$$ (as both a and c are negative) -------->(3)

So, from (1),(2) and (3),

$$c>a>d>b$$

So $$a,b,c,d$$ are now arranged in descending order.

Also, there are 4 terms in total.

So, median = average of $$3^{rd}$$ term and $$4^{th}$$ term = $$\dfrac{a+d}{2}$$