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Ответ:

$tga=\dfrac{3}{4}\\\\00\\\\1+tg^2a=\dfrac{1}{cos^2a}\ \ ,\ \ \dfrac{1}{cos^2a}=1+\dfrac{9}{16}=\dfrac{25}{16}\ \ \to \ \ cosa=\dfrac{4}{5}\\\\\\sina=\sqrt{1-cos^2a}=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5}\\\\\\sin2a=2\, sina\cdot cosa=2\cdot \dfrac{3}{5}\cdot \dfrac{4}{5}=\dfrac{24}{25}\\\\\\cos^3a-sin^3a=\dfrac{64}{125}-\dfrac{27}{125}=\dfrac{37}{125}$

$\dfrac{1+\frac{1}{2}sin2a}{cos^3a-sin^3a}=\dfrac{1+\frac{12}{25}}{\frac{37}{125}}=\dfrac{37\cdot 125}{25\cdot 37}=5$