Given
$$ y=c_1e^{-2t}\sin t+c_2y_2(t)+c_3y_3(t)\,. $$
That means, you've already had one of the fundamental solutions of the homogeneous ode. So, you can exploit it to find $a$ as a first step. Substituting $ y_1(x)=e^{-2 t}\sin(t) $ in the ode and simplifying you will find that $a=4$. Then the ode becomes
$$ y'''+5y''+9y'+5y=0 \Rightarrow (D^3+5D+9D^2+5)y= Ay=0 \,,$$
where
$$ A := D^3+5D+9D^2+5\,. $$
Now, the task is how to find the other two solutions. Recalling the annihilator method we used in the other problem, we have,
$$ y_1(x) = e^{-2 t}\sin(t) = \frac{1}{2i}e^{(-2+i)t} - \frac{1}{2i}e^{(-2-i)t} $$
Applying the annihilator $ (D-(-2+i))(D-(-2-i)) $ to the above equation gives
$$ (D-(-2+i))(D-(-2-i))y_1(t)= 0 \Rightarrow (D^2+4D+5)y_1= By_1 =0 \,, $$
where
$$ B = D^2+4D+5 $$
Now, if you divide $A$ by $B$ (division of polynomials), you get the other root of A. If you do that, you will get
$$A = (D+1)B= (D+1)(D-(-2+i))(D-(-2-i)) \,.$$
So, we were able to factor our operator. In fact, now you have the three roots you are looking for
$$ r_1 = -1 \,,\, r_2 = -2+i\,,\, r_3 = -2-i \,,$$
and the their corresponding solutions
$$ \left\{ y_1(t) = e^{-t} \,,\, y_2(t) = e^{(-2+i)t} \,,\, y_2(t) = e^{(-2-i)t}\right\}\,.$$
The general solution is given by
$$ y(x) = b_1 e^{-t} + b_2 e^{(-2+i)t} + b_3 e^{(-2-i)t} \,. $$
Now, manipulating the above general solution, exploiting the identity $e^{ix}=\cos(x)+i\sin(x)$, and then comparing with the general solution you were given
$$ y=c_1e^{-2t}\sin t+c_2y_2(t)+c_3y_3(t)\,, $$
you should be able to find $y_2(t)$ and $y_3(t)$. Solution is
$$ y_2(t) =e^{-t}\,,\, y_3(t)= e^{-2t}\cos(t) \,.$$