Number reduction before mod $$15\cdot25 \pmod {11}\equiv 4\cdot3 \pmod {11}$$
How does it work?
Full example
$$3^{(11-1)} \pmod{11} = 3\cdot27\cdot27\cdot27 \pmod{11}= 3\cdot5\cdot5\cdot5 \pmod{11} = 15\cdot25 \pmod{11}= 4\cdot3 \pmod{11} =1 \pmod{11}$$

$a\equiv b \pmod{11}$ means that $11$ divides $a-b$. So if we want to see that $$(a+11)(b+11)\equiv ab \pmod{11}$$ we should check that $11$ is a divisor of $$(a+11)(b+11)-ab=ab+11a+11b+11*11-ab=11a+11b+121$$ This is clearly the case (in your example we have that $a=4$ and $b=14$). You can do this again for $b$ and you see that $15\cdot 25 \equiv 3 \cdot 4 \pmod{11}$.
