Here is a systematic way to solve it,
$$ f_n - 3f_{n-1} - 12(-1)^n =0\,. $$
Shifting $n$ in the above equation gives
$$ f_{n+1} - 3f_{n} + 12(-1)^n =0\,. $$
add the two equations results in the homogeneous difference equation
$$ f_{n+1} -2 f_{n} - 3 f_{n-1} =0 $$
Now, we can solve the above recurrence relation, subject to the initial conditions $f_{0}=0$ and $f_2 = 12 $, which is equivalent to the original one. To solve it, just assume the has the form $f_n=r^n$ and substitute back into the difference equation and solve for $r$. Finding the roots of roots of the resulting polynomial in $r$ gives the solution
$$f_n = c_1 (-1)^n+c_2 3^n \,.$$
Exploiting the initial conditions gives
$$f_n = -3 (-1)^n+ 3^{n+1} \,.$$
Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence as:
$$
a_{n + 2} = - 2 a_{n + 1} + 15 a_n
$$
Running the recurrence "backwards" gives $a_0 = 6$ (starting at index 0 is nicer all around). Multiply the recurrence by $z^n$, sum over $n \ge 0$, and recognize:
\begin{align}
\sum_{n \ge 0} a_{n + 1} z^n &= \frac{A(z) - a_0}{z} \\
\sum_{n \ge 0} a_{n + 2} z^n &= \frac{A(z) - a_0 - a_1 z}{z^2}
\end{align}
to get:
$$
\frac{A(z) - 6 - 10 z}{z^2} = - 2 \frac{A(z) - 6}{z} + 15 A(z)
$$
Solving for $A(z)$, and writing as partial fractions, gives:
$$
A(z) = \frac{6 + 22 z}{1 + 2 z - 15 z^2}
= \frac{1}{1 + 5 z} + \frac{5}{1 - 3 z}
$$
Two geometric series:
$$
a_n = (-5)^n + 5 \cdot 3^n
$$
This gives $a_5 = -1910$.
Best Answer
One thing you can do is to consider the Sequence $$U(n)=\frac{T(n)}{(n-1)!}$$ Hence you can show $$ U(n+1)=U(n)+(n+1)(n+2) $$ You can then, sum this relation to find $U(n)$ (telescopic sums)