2. Let the sequence $\left\{a_{n}\right\}$ satisfy:
$$
a_{1}=1,4 a_{n+1}-a_{n+1} a_{n}+4 a_{n}=9 \text {. }
$$

Then $a_{2018}=$
2. $\frac{5}{3}$.

From the given, we have $\left(4-a_{n}\right)\left(4-a_{n+1}\right)=7$.
Let $b_{n}=4-a_{n}$. Then $b_{n} b_{n+1}=7$.
Also, $b_{1}=4-a_{1}=3$, so $b_{2}=\frac{7}{3}$.
In general, we have $b_{2 k-1}=3, b_{2 k}=\frac{7}{3}$.
Thus, $a_{2 k}=4-\frac{7}{3}=\frac{5}{3}, a_{2018}=\frac{5}{3}$.