Solution:

Using the identity
$$x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left((x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right)$$
we get
$$-3=a^{3}+b^{3}+2^{3}-6 a b=\frac{1}{2}(a+b+2)\left((a-b)^{2}+(a-2)^{2}+(b-2)^{2}\right)$$
Since $S=(a-b)^{2}+(a-2)^{2}+(b-2)^{2}$ must be positive, we conclude that $a+b+2<0$, i.e. that $a+b<-2$. Now $S$ can be bounded by
$$S \geqslant(a-2)^{2}+(b-2)^{2}=a^{2}+b^{2}-4(a+b)+8 \geqslant \frac{(a+b)^{2}}{2}-4(a+b)+8>18$$
Here, we have used the fact that $a+b<-2$, which we have proved earlier. Since $a+b+2$ is negative, it immediately implies that $a+b+2< -\frac{2 \cdot 3}{18} = -\frac{1}{3}$, i.e. $a+b< -\frac{7}{3}$ which we wanted.

Solution:

Writing $s=a+b$ and $p=a b$ we have
$$a^{3}+b^{3}-6 a b=(a+b)\left(a^{2}-a b+b^{2}\right)-6 a b=s\left(s^{2}-3 p\right)-6 p=s^{3}-3 p s-6 p$$
This gives $3 p(s+2)=s^{3}+11$. Thus $s \neq -2$ and using the fact that $s^{2} \geqslant 4 p$ we get
$$p=\frac{s^{3}+11}{3(s+2)} \leqslant \frac{s^{2}}{4}$$
If $s>-2$, then (1) gives $s^{3}-6 s^{2}+44 \leqslant 0$. This is impossible as
$$s^{3}-6 s^{2}+44=(s+2)(s-4)^{2}+8>0$$
So $s<-2$. Then from (1) we get $s^{3}-6 s^{2}+44 \geqslant 0$. If $s< -\frac{7}{3}$ this is again impossible as $s^{3}-6 s^{2}=s^{2}(s-6)< -\frac{49}{9} \cdot \frac{25}{3}<-44$. (Since $49 \cdot 25=1225>1188=44 \cdot 27$.) So $-\frac{7}{3}<s<-2$ as required.