In the rectangle, diagonal \(AC\) is the hypotenuse of right triangle \(ABC\).

By Pythagoras' theorem:

\(AC=\sqrt{6^2+8^2}=\sqrt{36+64}=10\).

For angle \(\angle BAC\), the opposite side is \(BC\) and the hypotenuse is \(AC\).

So, \(\sin \angle BAC=\dfrac{BC}{AC}=\dfrac{8}{10}=\dfrac{4}{5}\).

First find the other leg:

\(AC=\sqrt{15^2-9^2}=\sqrt{225-81}=12\).

For angle \(B\), the opposite side is \(AC\) and the hypotenuse is \(AB\).

So, \(\sin B=\dfrac{AC}{AB}=\dfrac{12}{15}=\dfrac{4}{5}\).

By the definition of sine:

\(\sin B=\dfrac{AC}{AB}=\dfrac{5}{13}\).

So \(AC=26\cdot\dfrac{5}{13}=10\text{ cm}\).

Now by Pythagoras' theorem:

\(BC=\sqrt{26^2-10^2}=\sqrt{676-100}=24\text{ cm}\).

In right triangle \(ABC\):

\(AC=\sqrt{24^2+7^2}=\sqrt{576+49}=25\).

For angle \(\angle BAC\), the adjacent leg is \(AB\) and the hypotenuse is \(AC\).

Therefore, \(\cos \angle BAC=\dfrac{AB}{AC}=\dfrac{24}{25}\).

Find the other leg:

\(BC=\sqrt{25^2-24^2}=\sqrt{625-576}=7\).

For angle \(B\), the adjacent leg is \(BC\) and the hypotenuse is \(AB\).

So, \(\cos B=\dfrac{BC}{AB}=\dfrac{7}{25}\).