Question 625305
Always remember, the inverse is a way to undo the equation.  
First problem: H(x)=1/3x+1.  Which is y=1/3x+1.  You subtract 1 from both sides, then you multiply both sides by three.  So the inverse is: 3(y-1).
H(x)=7/8x-2.  Which is y=7/8x-2.  First yo add 2 to both sides, then multiply by the reciprocal of 8/7.  You get 8/7(y+2). 
D(x)=5x+7/2x+4.  Combine like terms.  So, y=8.5x+4.  Subtract 4 from both sides, then divide by 8.5. So, (y-4)/(8.5).