Answer
$$\lim _{x \rightarrow a} [f(x)g(x)]=\frac{3}{4}$$
Work Step by Step
Since
$$
\begin{split}
\lim _{x \rightarrow a} f(x) &=\lim _{x \rightarrow a}\left(\frac{1}{2}[f(x)+g(x)]+\frac{1}{2}[f(x)-g(x)]\right) \\
& \frac{1}{2} \lim _{x \rightarrow a}[f(x)+g(x)]+\frac{1}{2} \lim _{x \rightarrow a}[f(x)-g(x)] \\
& =\frac{1}{2} \cdot 2+\frac{1}{2} \cdot 1\\
& =\frac{3}{2}
\end{split}
$$
and
$$
\begin{split}
\lim _{x \rightarrow a} g(x) &=\lim _{x \rightarrow a}([f(x)+g(x)]-f(x)) \\
&=\lim _{x \rightarrow a}[f(x)+g(x)]-\lim _{x \rightarrow a} f(x)\\
& =2-\frac{3}{2} \\
& =\frac{1}{2}.
\end{split}
$$
Then
$$
\begin{split}
\lim _{x \rightarrow a}[f(x) g(x)]&=\left[\lim _{x \rightarrow a} f(x)\right]\left[\lim _{x \rightarrow a} g(x)\right]\\
&=\frac{3}{2} \cdot \frac{1}{2}\\
&=\frac{3}{4}.
\end{split}
$$
Therefore, if $$\lim _{x \rightarrow a} [f(x)+g(x)]=2$$
and
$$\lim _{x \rightarrow a} [f(x)-g(x)]=1$$ ,
then
$$\lim _{x \rightarrow a} [f(x)g(x)]=\frac{3}{4}$$
