From the equivalent transformation III of $\S 1$, we know that the solution to this congruence equation is the same as the solution to the congruence equation
$$4\left(x^{2}+x+7\right) \equiv 0\left(\bmod 3^{3}\right)$$

This congruence equation is
$$(2 x+1)^{2}+27 \equiv(2 x+1)^{2} \equiv 0\left(\bmod 3^{3}\right)$$

It is evident that the solution to this congruence equation (referring to the value of $x$) is the same as the solution to
$$2 x+1 \equiv 0\left(\bmod 3^{2}\right)$$

Direct calculation shows that the solution is
$$x \equiv 4\left(\bmod 3^{2}\right) \text {. }$$

Therefore, the solution to the original congruence equation is (why)
$$x \equiv-5,4,13\left(\bmod 3^{3}\right),$$

The number of solutions is 3.