1. Use the given angle ranges

Since $0<\alpha,\beta<\dfrac{\pi}{4}$, we have:

• $0<\alpha+\beta<\dfrac{\pi}{2}$, so $\cos(\alpha+\beta)>0$
• $-\dfrac{\pi}{4}<\alpha-\beta<\dfrac{\pi}{4}$, and given $\sin(\alpha-\beta)=\dfrac{5}{13}>0$, hence $\alpha-\beta\in\left(0,\dfrac{\pi}{4}\right)$, so $\cos(\alpha-\beta)>0$

Thus we can determine both sine and cosine of these angles positively.

2. Find $\sin(\alpha+\beta)$ and $\cos(\alpha-\beta)$

Given $\cos(\alpha+\beta)=\frac{3}{5}$ so $\sin(\alpha+\beta)=\sqrt{1-\left(\frac{3}{5}\right)^2}=\frac{4}{5}$ (because $\alpha+\beta\in(0,\pi/2)$).

Also given $\sin(\alpha-\beta)=\frac{5}{13}$ so $\cos(\alpha-\beta)=\sqrt{1-\left(\frac{5}{13}\right)^2}=\frac{12}{13}$ (because $\alpha-\beta\in(0,\pi/4)$).

3. Use $2\alpha=(\alpha+\beta)+(\alpha-\beta)$

Let $X=\alpha+\beta,\qquad Y=\alpha-\beta$ Then $2\alpha=X+Y$ Hence, $\tan(2\alpha)=\tan(X+Y)=\frac{\tan X+\tan Y}{1-\tan X\tan Y}$

First compute: $\tan X=\frac{\sin X}{\cos X}=\frac{4/5}{3/5}=\frac{4}{3}$ $\tan Y=\frac{\sin Y}{\cos Y}=\frac{5/13}{12/13}=\frac{5}{12}$

Therefore, $\tan(2\alpha)=\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\cdot\frac{5}{12}}$

4. Simplify

Numerator: $\frac{4}{3}+\frac{5}{12}=\frac{16}{12}+\frac{5}{12}=\frac{21}{12}=\frac{7}{4}$

Denominator: $1-\frac{20}{36}=1-\frac{5}{9}=\frac{4}{9}$

So, $\tan(2\alpha)=\frac{7/4}{4/9}=\frac{7}{4}\cdot\frac{9}{4}=\frac{63}{16}$

5. Match with options

$\tan(2\alpha)=\frac{63}{16}$ So the correct option is D.