From \begin{align} \sin\alpha + \sin\beta=\frac{3}{5}\qquad(1)\\ \cos\alpha + \cos\beta=\frac{3}{5}\qquad (2)\\ \end{align} $(1)^2 + (2)^2 \implies (\sin\alpha+\sin\beta)^2 +(\cos\alpha+\cos\beta)^2 =(\frac{3}{5})^2 +(\frac{4}{5})^2=1$ It follows that $$2+2\sin\alpha\sin\beta + 2\cos\alpha\cos\beta=1\implies \cos(\alpha-\beta)=\boxed{-\frac{1}{2}}$$ Also, we have: $$(1)\implies 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-beta}{2}=\frac{3}{5}$$ $$(2)\implies 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-beta}{2}=\frac{4}{5}$$ Dividing these two equation leads to $\tan\frac{\alpha +\beta}{2} = \frac{3}{4}$ Hence $$\sin(\alpha+\beta)=\frac{2\tan\frac{\alpha+\beta}{2}}{1+\tan^2\frac{\alpha+\beta}{2}}=\boxed{\frac{24}{25}}$$