Notice that in a triangle, $A>B \Leftrightarrow \sin A>\sin B$ and $\cos A<\cos B$. Without loss of generality, assume $B$ is in the middle, then by the rearrangement inequality we get $\sin A \cos B+\sin B \cos C+\sin C \cos A \leqslant \sin A \cos C+\sin B \cos B+\sin C \cos A$ $=\sin B+\sin B \cos B=\sin B(1+\cos B)=\frac{\sqrt{3} \sin B(1+\cos B)}{\sqrt{3}} \leqslant \frac{(\sqrt{3} \sin B+1+\cos B)^{2}}{4 \sqrt{3}}$ $\leqslant \frac{(2+1)^{2}}{4 \sqrt{3}}=\frac{3 \sqrt{3}}{4}$, equality holds when $A=B=C=60^{\circ}$. Therefore, $\alpha=\frac{27}{16}$,
the sum of the numerator and denominator of the simplest fraction of $\alpha$ is 43.
2. Let the three interior angles of $\triangle A B C$ be $A, B, C$, and denote the maximum value of $(\sin A \cos B+\sin B \cos C+\sin C \cos A)^{2}$ as $\alpha$. Then the sum of the numerator and denominator of $\alpha$ when written as a simplest fraction is $\qquad$.