1. Depending on the real parameter $a$, solve the equation

$$
\sqrt{2 x-a}-\sqrt{x-1}=2
$$

in the set of real numbers.
1. We immediately have the conditions $x \geqslant 1$ and $2 x \geqslant a$. Transform the equation to $\sqrt{2 x-a}=2+\sqrt{x-1}$ and square it, after which we get $2 x-a=4+4 \sqrt{x-1}+x-1$, i.e., $x-(3+a)=4 \sqrt{x-1}$. After squaring again, with the condition $x \geqslant 3+a$, we get $x^{2}-2 x(3+a)+\left(9+6 a+a^{2}\right)=16(x-1)$, i.e., $x^{2}-(22+2 a) x+25+6 a+a^{2}=0$. Solving this quadratic equation, we get:

$x_{1,2}=\frac{22+2 a \pm \sqrt{\left(484+88 a+4 a^{2}\right)-\left(100+24 a+4 a^{2}\right)}}{2}=\frac{22+2 a \pm \sqrt{384+64 a}}{2}=11+a \pm 4 \sqrt{6+a}$

For $a-6$, check which solutions satisfy the conditions of definition $x \geqslant 1, 2 x \geqslant a$, and $x \geqslant 3+a$. The first condition reduces to $11+a \pm 4 \sqrt{6+a} \geqslant 1$, i.e., $10+a \geqslant \mp 4 \sqrt{6+a}$; after squaring, we get $100+20 a+a^{2} \geqslant 96+16 a$, i.e., $a^{2}+4 a+4 \geqslant 0$, which is always satisfied (since the left side is equal to $(a+2)^{2}$). The second condition reduces to $22+2 a \pm 8 \sqrt{6+a} \geqslant a$, i.e., $22+a \geqslant \mp 8 \sqrt{6+a}$; after squaring, we get $484+44 a+a^{2} \geqslant 384+64 a$, i.e., $a^{2}-20 a+100 \geqslant 0$, which is always satisfied (since the left side is equal to $(a-10)^{2}$). The third condition remains. It reduces to $11+a \pm 4 \sqrt{6+a} \geqslant 3+a$, i.e., $8 \pm 4 \sqrt{6+a} \geqslant 0$, and finally $2 \pm \sqrt{6+a} \geqslant 0$. It is always true that $2+\sqrt{6+a} \geqslant 0$, so we always have one solution. The inequality $2-\sqrt{6+a} \geqslant 0$ reduces (after rearranging and squaring) to $4 \geqslant 6+a$, i.e., $a \leqslant-2$, so then we have a second solution.

Thus, summarizing:

- for $a<-6$ there are no real solutions;
- for $a=-6$ the only solution is $x=5$;
- for $-6<a \leqslant-2$ we have two solutions, $x_{1 / 2}=11+a \pm 4 \sqrt{6+a}$;
- for $-2<a$ we have one solution, $x=11+a+4 \sqrt{6+a}$.