解：$(1)①$由勾股定理可得$AB=\sqrt {AC^2+BC^2}=10$
∴$sin A=\frac {BC}{AB}=cosB=\frac {6}{10}=\frac {3}{5}，$$cos A=\frac {AC}{AB}=sinB=\frac {8}{10}=\frac {4}{5}$
②由勾股定理可得$BC=\sqrt {AB^2-AC^2}=24$
∴$sin A=cos B=\frac {BC}{AB}=\frac {24}{25}，$$cos A=sin B=\frac {AC}{AB}=\frac {7}{25}$
③由勾股定理可得$AB=\sqrt {AC^2+BC^2}=\sqrt {29}$
∴$sin A=cos B=\frac {BC}{AB}=\frac 5{\sqrt {29}}=\frac {5\sqrt {29}}{29}，$$cos A=sin B=\frac {AC}{AB}=\frac 2{\sqrt {29}}=\frac {2\sqrt {29}}{29}$
$(2)$当$∠A+∠B=90°$时，$sin A=cos B，$$cosA=sinB$
