1. First, write the original equation as follows:

$\left((x-1)^{2}-2\right)^{2}+3(x-1)^{2}-16=0$.

Let $y=(x-1)^{2}$, then for $y$ we get the equation $y^{2}-y-12=0$, whose roots are 4 and -3. The real solutions are highlighted as follows: $(x-1)^{2}=$ 4.
Answer: $x_{1}=3, x_{2}=-1$.