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