(1) Method 1: Let the real root of the equation be $\alpha$, and the other root be $a+b \mathrm{i}(a, b \in \mathbf{R})$, then
$$
\begin{aligned}
x^{2}+(1-2 \mathrm{i}) x+3 m-\mathrm{i} & =(x-\alpha)[x-(a+b \mathrm{i})] \\
& =x^{2}-(\alpha+a+b \mathrm{i}) x+\alpha(a+b \mathrm{i}) \\
& =0 .
\end{aligned}
$$

By comparing coefficients, we get $b=2, \alpha \cdot b=-1, \alpha+a=-1$, so $\alpha=-\frac{1}{2}, a=-\frac{1}{2}$. Therefore, the other root is $-\frac{1}{2}+2 \mathrm{i}$. Hence, the answer is C.
Method 2: Let the roots be $\alpha, \beta$, where $\alpha$ is a real number, then
$$
\alpha^{2}+(1-2 \mathrm{i}) \alpha+3 m-\mathrm{i}=\left(\alpha^{2}+\alpha+3 m\right)-(2 \alpha+1) \mathrm{i}=0,
$$

which gives
$$
\left\{\begin{array}{l}
\alpha^{2}+\alpha+3 m=0 \\
2 \alpha+1=0
\end{array}\right.
$$

Solving these, we get $\alpha=-\frac{1}{2}$. Since
$$
\alpha+\beta=-(1-2 \mathrm{i}),
$$

we have
$$
\beta=-(1-2 \mathrm{i})-\alpha=-\frac{1}{2}+2 \mathrm{i} .
$$

Hence, the answer is C.
(1) If the equation $x^{2}+(1-2 \mathrm{i}) x+3 m-\mathrm{i}=0(m \in \mathbf{R})$ has one real root, then its other root is ( ).
(A) $-\frac{3}{2}+2 \mathrm{i}$
(B) $\frac{3}{2}-2 \mathrm{i}$
(C) $-\frac{1}{2}+2 \mathrm{i}$
(D) $\frac{1}{2}-2 \mathrm{i}$