Example 6 Let the quadratic function $f(x)=a x^{2}+b x+c$ have its graph symmetric about the $y$-axis. Given that $a+b=1$, and if the point $(x, y)$ is on the graph of $y=f(x)$, then the point $\left(x, y^{2}+1\right)$ is on the graph of the function $g(x)=f(f(x))$.
(1) Find the analytical expression for $g(x)$;
(2) Let $F(x)=g(x)-\lambda f(x)$. Does there exist a real number $\lambda$ such that $F(x)$ is a decreasing function in $\left(-\infty,-\frac{\sqrt{2}}{2}\right]$ and an increasing function in $\left(-\frac{\sqrt{2}}{2}, 0\right)$?
Solution (1) Since the axis of symmetry of $f(x)=a x^{2}+b x+c$ is the $y$-axis, i.e., $-\frac{b}{2 a}=0$, hence $b=0$, thus $a=1$, $f(x)=x^{2}+c$. Let $\left(x_{0}, y_{0}\right)$ be on the graph of $y=f(x)$, by the problem, the point $\left(x_{0}, y_{0}^{2}+1\right)$ is on the graph of $y=f(f(x))$, i.e., $y_{0}=x_{0}^{2}+c, y_{0}^{2}+1=\left(x_{0}^{2}+c\right)^{2}+c$. From the above two equations, it is easy to know that $c=1$. Therefore, $f(x)=x^{2}+1$. Hence, $g(x)=\left(x^{2}+1\right)^{2}+1$.
(2) From (1), we get $F(x)=g(x)-\lambda f(x)=x^{4}+(2-\lambda) x^{2}+2-\lambda$. Let $x_{1}<0$, but $x_{1}^{2}-x_{2}^{2}>0$, so it is sufficient that $x_{1}^{2}+x_{2}^{2}+2-\lambda>0$, thus $\lambda<3$, therefore as long as $\lambda \leqslant 3, F(x)$ is a decreasing function in $\left(-\infty,-\frac{\sqrt{2}}{2}\right)$. Similarly, when $\lambda \geqslant 3$, $F(x)$ is an increasing function in $\left(-\frac{\sqrt{2}}{2}, 0\right)$. In summary, there exists a unique real number $\lambda=3$, such that the corresponding $F(x)$ satisfies the conditions.