$12.55 \int_{0}^{\pi / 4}(\sin 2 t-\cos 2 t)^{2} d t$.
12.55 Raising the expression $\sin 2 t-\cos 2 t$ to the square and using known trigonometric formulas, we find

$$
\begin{aligned}
& \int_{0}^{\pi / 4}(\sin 2 t-\cos 2 t)^{2} d t= \\
& =\int_{0}^{\pi / 4}\left(\sin ^{2} 2 t-2 \sin 2 t \cos 2 t+\cos ^{2} 2 t\right) d t= \\
& =\int_{0}^{\pi / 4} d t-\int_{0}^{\pi / 4} \sin 4 t d t=\left.\left(t+\frac{1}{4} \cos 4 t\right)\right|_{0} ^{\pi / 4}=\frac{\pi}{4}-\frac{1}{4}-\frac{1}{4}=\frac{\pi-2}{4}
\end{aligned}
$$

Answer: $0.25(\pi-2)$.