2. Find all values of the real parameter \(a\) for which some two distinct solutions of the equation

\[
x^{4} - a x^{3} + x^{2} + a = 0
\]

(in the set \(\mathbb{C}\)) have a sum equal to 1.
2. Let $y, z, u$ and $v$ be the roots of the given polynomial, where $y+z=1$. From Vieta's formulas, we have $y+z+u+v=a$, hence $u+v=a-1$; further, $(y+z)u + (y+z)v + uv + yz = 1$, i.e., $u+v + uv + yz = 1$, and this, together with the previous conclusion, gives $uv + yz = 1 - (u+v) = 1 - (a-1) = 2 - a$; and then, $(y+z)uv + (u+v)yz = 0$, i.e., $uv + (a-1)yz = 0$, and subtracting the previously obtained equality from this one yields $(a-2)yz = a-2$, i.e., $(a-2)(yz - 1) = 0$.

If $yz = 1$, then from the previous equalities we get $uv = 1 - a$, and for the last Vieta's formula to be satisfied, $yzuv = a$, it must be that $1 - a = a$, i.e., $a = \frac{1}{2}$. This value is a solution to the problem because the numbers $y, z, u$, and $v$ satisfy all four Vieta's formulas for the considered polynomial, and thus they must indeed be its roots.

In the case $a = 2$, the equation from the problem reduces to $x^4 - 2x^3 + x^2 + 2 = 0$, i.e., $0 = (x^2 - x)^2 + 2 = (x^2 - x - 2i)(x^2 - x + 2i)$. From Vieta's formulas applied to the polynomial in the first parenthesis (or the second), it follows that the sum of the roots of that polynomial is 1, so $a = 2$ is also a possible value.

Thus, the problem has two solutions: $a \in \left\{\frac{1}{2}, 2\right\}$.