12. (13 points) In a game activity of a TV entertainment program, each participant needs to complete three tasks: $A$, $B$, and $C$. It is known that the probabilities for participant Jia to complete tasks $A$, $B$, and $C$ are $\frac{3}{4}$, $\frac{3}{4}$, and $\frac{2}{3}$, respectively. Each task is independent of the others.
(1) Participant Jia attempts tasks $A$, $B$, and $C$ once each. Find the probability that Jia completes at least one of the tasks.
(2) The activity requires task $A$ to be done twice, task $B$ to be done twice, and task $C$ to be done three times. If both attempts of task $A$ are successful, then task $B$ is attempted, and the participant earns points $a$; if both attempts of task $B$ are successful, then task $C$ is attempted, and the participant earns points $3a$; if at least two out of the three attempts of task $C$ are successful, the participant successfully completes the challenge and earns points $6a$ (points are not cumulative), and each task is independent of the others. Let $X$ represent the number of points participant Jia earns. Write the distribution of $X$ and find its expected value.
12. (1) Let the events of contestant A completing projects $A$, $B$, and $C$ be denoted as events $A$, $B$, and $C$, respectively, and assume they are mutually independent. The event of completing at least one project is denoted as event $D$.
$$
\begin{array}{l}
\text { Then } P(D)=1-P(\bar{A} \bar{B} \bar{C}) \\
=1-\frac{1}{4} \times \frac{1}{4} \times \frac{1}{3}=\frac{47}{48} .
\end{array}
$$
(2) The values of $X$ are $0$, $a$, $3a$, and $6a$. Then
$$
\begin{array}{l}
P(X=0)=\frac{1}{4}+\frac{3}{4} \times \frac{1}{4}=\frac{7}{16}, \\
P(X=a)=\left(\frac{3}{4}\right)^{2}\left(\frac{1}{4}+\frac{3}{4} \times \frac{1}{4}\right)=\frac{63}{256}, \\
P(X=3a) \\
=\left(\frac{3}{4}\right)^{2}\left(\frac{3}{4}\right)^{2}\left(\left(\frac{1}{3}\right)^{2}+\mathrm{C}_{2}^{1}\left(\frac{1}{3}\right)^{2} \frac{2}{3}\right)=\frac{21}{256}, \\
P(X=6a)=1-\frac{7}{16}-\frac{63}{256}-\frac{21}{256}=\frac{60}{256} .
\end{array}
$$

Thus, the distribution of $X$ is as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|}
\hline$X$ & 0 & $a$ & $3a$ & $6a$ \\
\hline$P$ & $\frac{7}{16}$ & $\frac{63}{256}$ & $\frac{21}{256}$ & $\frac{60}{256}$ \\
\hline
\end{tabular}

Therefore, $\mathrm{E}(X)$
$$
\begin{array}{l}
=0 \times \frac{7}{16}+a \times \frac{63}{256}+3a \times \frac{21}{256}+6a \times \frac{60}{256} \\
=\frac{243}{128} a .
\end{array}
$$