\[\begin{array}{l}
f(x) = \frac{{x - 1}}{{2{x^2} - 3x + 1}}\\
f( - 2) = \frac{{( - 2) - 1}}{{2.{{( - 2)}^2} - 3( - 2) + 1}} = \frac{{ - 3}}{{15}} = \frac{{ - 1}}{5}\\
f(0) = \frac{{0 - 1}}{{{{2.0}^2} - 0.x + 1}} = \frac{{ - 1}}{1} = - 1\\
f(2) = \frac{{2 - 1}}{{{{2.2}^2} - 3.2 + 1}} = \frac{1}{3}\\
f(3) = \frac{{3 - 1}}{{{{2.3}^2} - 3.3 + 1}} = \frac{2}{{10}} = \frac{1}{5}\\
f(\sqrt 2 ) = \frac{{\sqrt 2 - 1}}{{2{{(\sqrt 2 )}^2} - 3(\sqrt 2 ) + 1}} = \frac{{\sqrt 2 - 1}}{{5 - 3\sqrt 2 }} = \frac{{1 + 2\sqrt 2 }}{7}\\
f(x) = 2|x - 1| + 3|x| - 2\\
f( - 2) = 2.| - 2 - 1| + 3| - 2| - 2 = 2.3 + 3.2 - 2 = 10\\
f(0) = |0 - 1| + 3|0| - 2 = 1 + 0 - 2 = - 1\\
f(2) = |2 - 1| + 3|2| - 2 = 1 + 3.2 - 2 = 5\\
f(3) = |3 - 1| + 3|3| - 2 = 2 + 9 - 2 = 9\\
f(\frac{1}{2}) = |\frac{1}{2} - 1| + 3|\frac{1}{2}| - 2 = \frac{1}{2} + \frac{3}{2} - 2 = 0\\
f(\sqrt 3 ) = |\sqrt 3 - 1| + 3|\sqrt 3 | - 2 = \sqrt 3 - 1 + 3\sqrt 3 - 2 = 4\sqrt 3 - 3(do:\sqrt 3 - 1 > 0)\\
f(1 + \sqrt 2 ) = |1 + \sqrt 2 - 1| + 3|1 + \sqrt 2 | - 2\\
= \sqrt 2 + 3(1 + \sqrt 2 ) - 2(do:\sqrt 2 + 1 > 0)\\
= \sqrt 2 + 3 + 3\sqrt 2 - 2\\
= 4\sqrt 2 + 1
\end{array}\]