(2) Hint: From the given, we know $f^{\prime}(x)=3 x^{2}+2 a x+1$, and $x=-\frac{1}{3}$ is an extremum point of the function $f(x)$, i.e.,
$$
f^{\prime}\left(-\frac{1}{3}\right)=-\frac{2}{3} a+\frac{4}{3}=0,
$$

Solving for $a$ yields $a=2$.
(1) Given the function $f(x)=x^{3}+a x^{2}+x+1(a \in \mathbf{R})$ is decreasing in the interval $\left(-\frac{2}{3},-\frac{1}{3}\right)$ and increasing in the interval $\left(-\frac{1}{3},+\infty\right)$, then $a=$