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