Is this system time variant?
This system is time variant because plugging in \$x(t-a)\$ does not equal \$y(t-a)\$. 

For the first case you get:
\$y_1(t)=sin(t)x(t-a-2)\$

However, if you offset the output by a you get:
 \$y_2(t)=y(t-a)=sin(t-a)x(t-a-2)\$ 

Since \$y_1(t)\$ does not equal \$y_2(t)\$ the system is time variant. Look at the first example on this page if you are confused: https://en.wikipedia.org/wiki/Time-invariant_system