How to construct a triangle with $BC=7.5$ cm. $\angle ABC$=$60$° and $AC-AB=1.5$ cm.
At first I constructed $BC$ then $\angle ABC$ ,but I don't know what to do next. Please help me.
Using the cosine rule is not the way to solve this problem simply and efficiently. This is a problem about construciton, not trigonometry. You are not supposed to calculate values that are not given.
Suppose that triangle $ABC$ is the solution. Draw a circular arc $l$ with center at point $A$ and radius $AC$ until it meets the ray $AB$ in point $C'$. Obviously $BC'$=$AC-AB$, which is given. So it is possible to construct triangle $BCC'$: we know $BC$, $BC'$ and $\angle CBC'=180^\circ-\angle ABC=120^\circ$.
Triangle $ACC'$ is isosceles so the point $A$ has to be on the median $n$ of segment $CC'$. After the construciton of triangle $CBC'$ just extend $C'B$ until it meets the median of $CC'$. The intersection point is actually your point $A$.
Problems like this one do not need trigonometry at all.