$解: (1)构建一个直角三角形ABC$
$当AB=3,AC=4$
$∴cosA =\sqrt{AC²-AB²}=\sqrt{7}$
$tan α= tan A=\frac {BC}{AB}=\frac {\sqrt{7}}{3}$
$∵\sqrt{7}\lt 3$
$∴\frac {\sqrt{7}}{3}<1<\frac {5}{4}$
$∴tan α\lt tan β$
$∴α\lt β$
$(2)构建直角三角形ABC。$
$当BC=4,AC= 10时$
$sinA=\frac {BC}{AC}= 0.4$
$AB=\sqrt{AC²-BC²}= 2\sqrt{21}$
$cos α= cos-A= \frac {AB}{BC}=\frac {\sqrt{21}}{5}≈0.92$
$∴cosa>cosβ$
$∴a\lt β$