First thing I have done is choosing $y=[x-\frac{\pi}{4}]+=0+$
$\lim_{y \rightarrow 0+}[tg(\frac{3\pi}{8}+y)]^{tg(2y+\frac{\pi}{2})}$
I know that $tg(\frac{3\pi}{8})=ctg(\frac{\pi}{8})$, $ctg(\frac{\alpha}{2})=\frac{1+cos(\alpha)}{sin(\alpha)}$ and $tg(y) \sim y$
I'm not sure about $tg(\frac{\pi}{2}+0)\rightarrow -\infty$ when $y \rightarrow0+$
I have tried any kind of manipulations with these identities and $tg(\alpha+\beta)$, but couldnt find the solution.
Note that $\tan \frac{3\pi}{8} = 1 + \sqrt{2} > 1$. Thus you don't have an indeterminate form. For $\frac{\pi}{4} < x < \frac{3\pi}{8}$, we have $\tan (2x) < 0$ and $1 + \sqrt{2} < \tan \left(\frac{\pi}{8} + x\right)$ and thus
$$0 < \Bigl(\tan \bigl(\tfrac{\pi}{8} + x\bigr)\Bigr)^{\tan (2x)} < (1+\sqrt{2})^{\tan (2x)}.$$
Now $\lim\limits_{x\searrow \frac{\pi}{4}} \tan (2x) = -\infty$, and hence by squeezing, we obtain
$$0 \leqslant \lim_{x\searrow \frac{\pi}{4}} \Bigl(\tan \bigl(\tfrac{\pi}{8} + x\bigr)\Bigr)^{\tan (2x)} \leqslant \lim_{x\searrow \frac{\pi}{4}} (1+\sqrt{2})^{\tan (2x)} = \lim_{\alpha \to -\infty} (1+\sqrt{2})^\alpha = 0.$$