$解：(1)在Rt△BMN中，∵BN=3，MN=4$
$∴BM=\sqrt {BN^2+MN^2}=5$
$∴sin B=\frac {MN}{BM}=\frac 45，cos B=\frac {BN}{BM}=\frac 35，tan B=\frac {MN}{BN}=\frac 43$
$(2)∵MN⊥AB$
$∴∠MNB=∠C=90°$
$∴∠B+∠BMN=∠B+∠A=90°$
$∴∠BMN=∠A$
$在Rt△ABC中，∵AB=10，BC=5$
$∴AC=\sqrt {AB^2-BC^2}=5\sqrt 3$
$∴sin ∠BMN=sin A=\frac {BC}{AB}=\frac 12，cos ∠BMN=cosA=\frac {AC}{AB}=\frac {\sqrt 3}2$
$tan ∠BMN=tan A=\frac {BC}{AC}=\frac {\sqrt 3}3$
$(3)∵cos ∠BMN=cosA=\frac {AC}{AB}=\frac 34$
$不妨设AC=3x，AB=4x$
$在Rt△ABC中，∵AC=3x，AB=4x$
$∴BC=\sqrt {AB^2-AC^2}=\sqrt 7x$
$∴sinB=\frac {AC}{AB}=\frac {3x}{4x}=\frac 34，sinA=\frac {BC}{AB}=\frac {\sqrt 7x}{4x}=\frac {\sqrt 7}4，tanA=\frac {BC}{AC}=\frac {\sqrt 7x}{3x}=\frac {\sqrt 7}3$
