253. Given three statements:
1) the equation $x^{2}-a x+1=0$ has no roots (real);
2) the equality $|a-2|=2-a$ holds;
3) the system of equations

$$
\left\{\begin{array}{l}
x+y^{2}=a \\
x-\sin ^{2} y=-3
\end{array}\right.
$$

has a unique solution.

Find all values of $a$ for which two of these statements are true and one is false.
$\triangle$ The first statement is true if $a^{2}-4<0$, i.e., when $-2<a<2$. The second statement is true if $a \leq 2$.

To determine when the third statement is true, note that based on the form of the given system of equations, if ( $x_{0} ; y_{0}$ ) is a solution to the system, then ( $x_{0} ;-y_{0}$ ) is also a solution. Since the system has a unique solution, these solutions must coincide. Therefore, $y_{0}=-y_{0}$, which implies $y_{0}=0$. Substituting $y=0$ into the system, we find $x$: from the first equation, $x=a$, and from the second equation, $x=-3$. Thus, $a=-3$.

Now, let's verify if the converse statement is true: if $a=-3$, does the system of equations have a unique solution? When $a=-3$, the system becomes

$$
\left\{\begin{array}{l}
x+y^{2}=-3 \\
x-\sin ^{2} y=-3
\end{array}\right.
$$

Then $x+y^{2}=x-\sin ^{2} y$, which simplifies to $y^{2}+\sin ^{2} y=0$. Therefore, $y=0$, and from this, $x=-3$.

Thus, the third statement is true only when $a=-3$.

Now, let's determine for which values of $a$ exactly two of the three given statements are true.

The first and second statements are true on the intersection (common part) of the intervals $(-2 ; 2)$ and ( $-\infty ; 2]$, i.e., on the interval $(-2 ; 2)$. In this case, the third statement is false.

The second and third statements are true only when $a=-3$. In this case, the first statement is false.

Finally, the first and third statements cannot be true at the same values of $a$.

Answer: $-3,(-2 ; 2)$.