(a) We get that $f(x) = x^2 - 2x + 2$, we let $g(x) = f(x) - x$, thus $g(x) = x^2 - 3x + 2 = (x - 2)(x - 1)$. Thus set of real roots of $g(x)$ is $\{1 , 2\}$. We let $h(x) = f(f(x)) - x $, thus $h(x) = f(x^2 - 2x + 2) - x = (x^2 - 2x + 2)^2 - 2(x^2 - 2x + 2) + 2 - x = x^4 - 4x^3 + 6x^2 - 5x + 2$. Now we check whether $0$ or $\pm 1$ are roots of $h(x)$. We find that $1$ is root of $h(x)$. Thus $h(x) = (x - 1)(x - \alpha)(x - \beta)(x - \gamma)$, where $\alpha$,$\beta$,$\gamma$ are roots of $h(x)$. Performing polynomial division we get that $\frac{h(x)}{x - 1} = x^3 - 3x^2 + 3x - 2 = (x^3 - 3x^2 + 3x - 1) - 1 = (x - 1)^3 - 1 = (x - 2)((x - 1)^2 + (x - 1) + 1) = (x - 2)(x^2 - x + 1)$ , so $h(x) = (x - 1)(x - 2)(x^2 - x + 1)$, now note that $x^2 - x + 1$ has negative discriminant, thus $x^2 - x + 1$ has no real root and consequently set of real roots of $h(x)$ is $\{1,2\}$.

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