Answer
$\dfrac{3}{4}$
Work Step by Step
Let's use the limit property and expand the brackets
$\lim\limits_{x \to a}(f(x)+g(x))=\lim\limits_{x \to a}f(x)+\lim\limits_{x \to a}g(x)=f(a)+g(a)=2;$
At the same time
$\lim\limits_{x \to a}(f(x)-g(x))=\lim\limits_{x \to a}f(x)\lim\limits_{x \to a}g(x)=f(a)-g(a)=1;$
We have a system of equations
$$ \begin{cases}
f(a)+g(a)=2\\
f(a)-g(a)=1\\
\end{cases} $$
Let's add two equations
$f(a)+g(a)+f(a)-g(a)=2+1$
$2f(a)=3$ or $f(a)=\frac{3}{2}$
Let's subtract the second equations from the first
$f(a)+g(a)-(f(a)-g(a))=2-1$
$f(a)+g(a)-f(a)+g(a)=1$
$2g(a)=1$ or $g(a)=\frac{1}{2}$
Now it's easy to find the answer
$\lim\limits_{x \to a}(f(x)g(x))=\lim\limits_{x \to a}f(x)\lim\limits_{x \to a}g(x)=f(a)g(a)=\frac{3}{2}\cdot\frac{1}{2}=\frac{3}{4}$
