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I can solve it easily if I assume that $ 0 < \alpha, \beta < \frac{\pi}{2}$

But there is no mention of the quadrants in which $ \alpha $ and $ \beta $ lie in.

Is the question wrong ?

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    You know $\cos \alpha$ and $\sin \beta$ up to a sign. Look at the formulas you have for $\cos \frac{\alpha-\beta}{2}$ and see how they change when you vary the signs of those two quantities. Best case, you see that flipping each sign does nothing, worst case, you have four cases to consider, one for each possible combination of signs. – 2017-01-13
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    I did just that, and I got 4 different answers, only one of which matches $ \frac{8}{\sqrt{65}} $ – 2017-01-13
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    You're right: you can't prove that equality. – 2017-01-13
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    In that case, without verifying your work, I would say yes, the question is wrong, or at least incomplete, if it doesn't say something else that would give you a way of specifying those signs. – 2017-01-13

2 Answers 2

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Hint:

$\sin \alpha>0$ then $0<\alpha<\pi$ and $\cos \beta>0$ then $-\frac{\pi}2<\beta<\frac{\pi}2$ and $-\frac{\pi}2<-\beta<\frac{\pi}2$

Thus $$-\frac{\pi}2<\alpha-\beta<\frac{3\pi}2$$ and $$-\frac{\pi}4<\frac{\alpha-\beta}2<\frac{3\pi}4$$

1) $\sin^2\alpha+\cos^2\alpha=1$, then $\cos\alpha=\pm\sqrt{1-\sin^2\alpha}$ and $\sin \beta=\pm\sqrt{1-\cos^2\beta}$

2) $\cos (\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$

3)$|\cos \frac{\alpha-\beta}2|=\sqrt{\frac{1+\cos(\alpha-\beta)}2}$

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    @JonathanY. since $0<\alpha,\beta<\pi/2$ we have that $-\pi/2<(\alpha-\beta)/2<\pi/2$ hence the $\cos(\alpha-\beta)/2$ is nonnegative. – 2017-01-13
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    @JonathanY. my bad I did not read the statement properly. – 2017-01-13
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    Taking into account recent edits, if $\cos(\alpha-\beta)$ could be uniquely determined, then so would $\cos\frac{\alpha-\beta}{2}$. But *can* step 2 be completed like you wish? Also, consider that comments by OP suggest that he's tried this already. (In short, I'm asking if you followed through on your hints and have a complete solution in mind.) – 2017-01-13
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HINT:

\begin{align} \cos(\alpha/2-\beta/2)&=\sqrt{\frac{1+\cos(\alpha-\beta)}{2}}=\sqrt{(1+\cos\alpha\cos\beta+\sin\alpha\sin\beta)/2}\\ \end{align}

$$ \cos\alpha =\frac35, \cos\beta = \frac{5}{13},\, \sin\alpha=\frac45,\, \sin\beta =\frac{12}{13}, $$

plug in to get $ 8/\sqrt{65}$

For other quadrants calculate other three values also which are possible with inverse trig functions.

$$\sqrt{(1\pm\cos\alpha\cos\beta\pm\sin\alpha\sin\beta)/2}$$