For differentiability at \(x=1\): continuity gives \(-3a-2=a^2+b\) and equal derivatives give \(-6a=b\). From these: \(-3a-2=a^2-6a\Rightarrow a^2-3a+2=0\Rightarrow(a-1)(a-2)=0\). Since \(a>1\), we get \(a=2,\ b=-12\). So \(f(x)=\begin{cases}-6x^2-2,&x<1\\4-12x,&x\geq1\end{cases}\). Finding where \(f(x)=-20\): For \(x<1\): \(-6x^2-2=-20\Rightarrow x^2=3\Rightarrow x=-\sqrt{3}\) (taking the left root). For \(x\geq1\): \(4-12x=-20\Rightarrow x=2\). The enclosed area \(=\int_{-\sqrt{3}}^{1}[(-6x^2-2)-(-20)]\,dx+\int_{1}^{2}[(4-12x)-(-20)]\,dx\). First integral: \(\int_{-\sqrt{3}}^{1}(18-6x^2)\,dx=[18x-2x^3]_{-\sqrt{3}}^{1}=(18-2)-(-18\sqrt{3}+2\cdot3\sqrt{3})=(16)-(-18\sqrt{3}+6\sqrt{3})=16+12\sqrt{3}\). Second integral: \(\int_{1}^{2}(24-12x)\,dx=[24x-6x^2]_{1}^{2}=(48-24)-(24-6)=24-18=6\). Total area \(=22+12\sqrt{3}\). Thus \(\alpha=22,\beta=12\) and \(\alpha+\beta=34\).