Answer
$2$, not continuous.
Work Step by Step
Step 1. Simplify the function $f(x)=\frac{sin2x-sin(x)}{3x}=\frac{2sin(x)cos(x)-sin(x)}{3x}=\frac{sin(x)(2cos(x)-1)}{3x}$
Step 2. As $\lim_{x\to0}\frac{sin(x)}{x}=1$ and $\lim_{x\to0}\frac{2cos(x)-1}{3}=\frac{2-1}{3}=\frac{1}{3}$, we have $\lim_{x\to0}sec(\pi\frac{sin2x-sin(x)}{3x})=sec\frac{\pi}{3}=2$
Step 3. The function is not continuous at $x=0$ as it is not defined at this point (dividing by zero).