∵0＜α＜π/2
∴π/3＜α+π/3 ＜5π/6
又∵cos(π/3 +α)=-3/5 ＜ 0
∴π/2＜α+π/3 ＜5π/6
sin(π/3 +α)=4/5
∵π/2＜β＜π
∴-π/3＜2π/3 -β＜π/6
又∵sin(2π/3 -β)=5/13 ＞ 0
∴0＜2π/3 -β＜π/6
cos(2π/3 -β)=12/13
cos(β-α)=-cos(π-β+α)=-cos[(2π/3 -β)+(π/3 +α)]
=-[cos[(2π/3 -β)cos(π/3 +α) - sin(2π/3 -β)sin(π/3 +α)]
=-[12/13 * (-3/5) - 5/13 * 4/5]
=56/65