What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$ 
What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$

We have to make this function defined, by change the cosine into sine or tan. So,
$$\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}\right) = \lim_{t \to 0} \left(\frac{a}{t^2} - \frac{6}{t^2}\right) = -18$$
But I am confused what should I do next.

There's an error in your computation, and you should find first a Laurent expansion:
$$\frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}=\frac 1{t^2}\biggl(at-2\frac{\tan 3t}t\biggr)$$
Now the Taylor expansion of $\tan x$ at order $3$ is $\;\tan x=x+\dfrac{x^3}{3}+o\bigl(x^3\bigr)$, so
$$
\frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}=\frac 1{t^2}\biggl(a-\frac 2t\Bigl(3t+9t^3+ o\bigl(t^3\bigr)\Bigr)\biggr) =\frac{a-6}{t^2} -18+o(1).
$$
We conclude there's a limit at $0$ when $a=6$, and in this case the limit is indeed $-18$.
