2. For which real numbers $a$ is the minimum value of the function

$$
f(x)=4 x^{2}-4 a x+a^{2}-2 a+2
$$

on the interval $[0,2]$ equal to 3?
2. The minimum value of the function $f(x)$ on the set of real numbers is attained at $x=\frac{a}{2}$, (this is the vertex of the parabola). We need to consider three cases:
1) For $\frac{a}{2} \in [0,2]$ i.e., $a \in [0,4]$, it should be

$$
f\left(\frac{a}{2}\right)=4 \cdot\left(\frac{a}{2}\right)^{2}-4 a \cdot \frac{a}{2}+a^{2}-2 a+2=3
$$

from which it follows that $a=-\frac{1}{2} \notin [0,4]$, which is not possible.

7 points
2) For $\frac{a}{2} > 2$, i.e., $a > 4$, the function $f(x)$ is decreasing on the interval $[0,2]$, so it attains its minimum value at $x=2$. In this case,

$$
f(2)=16-8 a+a^{2}-2 a+2=3, \text{ i.e., } a^{2}-10 a+15=0
$$

From this, we get $a_{1,2}=5 \pm \sqrt{10}$. Since $a > 4$, we have $a=5+\sqrt{10}$.

Thus, the conditions of the problem are satisfied for $a=1-\sqrt{2}$ and $a=5+\sqrt{10}$. $\quad 7$ points