Prove that $\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B)$

You got off to a good start: $$ \sin(A+B)\sin(A-B) = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B)-\cos(A)\sin(B)) $$ This is of the form $(x+y)(x-y)$ so $$ \sin(A+B)\sin(A-B) = \sin^2(A)\cos^2(B)-\cos^2(A)\sin^2(B) $$ Eliminate the cosines (since $\sin^2(x)+\cos^2(x)=1$, so $\cos^2(x)=1-\sin^2(x)$) and expand: $$ \begin{align} \sin(A+B)\sin(A-B) &= \sin^2(A)(1-\sin^2(B))-(1-\sin^2(A))\sin^2(B)\\ &= \sin^2(A)-\sin^2(A)\sin^2(B)-\sin^2(B)+\sin^2(A)\sin^2(B)\\ &=\sin^2(A)-\sin^2(B) \end{align} $$ as required.

$x^2-y^2=(x+y)(x-y)$, we have $$\sin^2A-\sin^2B\\ =(\sin A+\sin B)(\sin A-\sin B)\\ =2\sin\frac{A+B}2\cos\frac{A-B}2\cdot2\cos\frac{A+B}2\sin\frac{A-B}2\\ =2\sin\frac{A+B}2\cos\frac{A+B}2\cdot2\sin\frac{A-B}2\cos\frac{A-B}2\\ =\sin(A+B)\sin(A-B)$$

You wrote: $$ \begin{align} & \phantom{={}}\sin^2(A) - \sin^2(B) \overset{(1)}{=} \sin(A + B)\sin(A -B) \\[8pt] & = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B) - \cos(A)\sin(B)) \\[8pt] & \overset{(2)}{=} (\sin(A)+\sin(B))(\cos(B)+\cos(A))(\sin(A)-\sin(B))(\cos(B)-\cos(A)) \\[8pt] & \overset{(3)}{=} (\sin(A)+\sin(B))^2(\cos(B)-\cos(A))^2 \end{align} $$ I've put numbers above things that are dubious or wrong.

The problem with the $(1)$ is that that is something you want to prove, not something you already know.

The problem with $(2)$ is that if you expand that product of four sums, you get $16$ terms, and if you expand the thing before it, you get a sum of only $8$ terms. You haven't shown that they're equal.

The problem with $(3)$ is that if you multiply $\cos B+\cos A$ by $\cos B-\cos A$, you get $\cos^2 B-\cos^2 A$, not $(\cos B-\cos A)^2$, and a similar thing applies to the sines.

I think some others have posted some things you can try instead.