已知 $f(x)=2x^3+ax^2+bx-10$（其中 $a$ 和 $b$ 为常数）含有因式 $(x-1)^2$。

1. State the values of $a$ and $b$ from part (a) and write the expression for $g(x)$

From part (a), $a = -14$ and $b = 22$.
$f(x) = 2x^3 - 14x^2 + 22x - 10$
$g(x) = 2x^3 - 14x^2 + 22x - 10 + k$

2. Apply the Factor Theorem to find $k$

Since $g(x)$ is divisible by $2x - 1$, by the Factor Theorem, $g\left(\frac{1}{2}\right) = 0$.
$2\left(\frac{1}{2}\right)^3 - 14\left(\frac{1}{2}\right)^2 + 22\left(\frac{1}{2}\right) - 10 + k = 0$
$2\left(\frac{1}{8}\right) - 14\left(\frac{1}{4}\right) + 11 - 10 + k = 0$
$\frac{1}{4} - \frac{7}{2} + 1 + k = 0$
$0.25 - 3.5 + 1 + k = 0$
$-2.25 + k = 0$
$k = 2.25$