It's definite so outcomes definite

$\begin{aligned} I_1 & = \int_0^\frac \pi 2 5\cos^2 x + 3 \sin^2 x \ dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\frac \pi 2 5\cos^2 x + 3\sin^2 x + 5\sin^2 x + 3 \cos^2 x \ dx \\ & = 4 \int_0^\frac \pi 2 1\ dx = 2 \pi \end{aligned}$

$\begin{aligned} I_2 & = \int_0^\frac \pi 2 \sin x \cos x \sqrt{\color{#3D99F6}25\sin^2 x + 9 \cos^2 x} \ dx & \small \color{#3D99F6} \text{Let }u= 25\cos^2 x + 9 \sin^2 x, \ du = 32\sin x \cos x \ dx \\ & = \int_9^{25} \frac {\sqrt u}{32} \ du \\ & = \frac {u\sqrt u}{48}\bigg|_9^{25} \\ & = \frac {49}{24} \end{aligned}$

$\implies \dfrac {I_1}{I_2} = \dfrac {2\pi}{\frac {49}{24}} = \boxed{\dfrac {48\pi}{49}}$

The numerator can be solved as

$\begin{aligned} I_1 = \displaystyle \int_0^{{\pi}/{2}} \left( 5 \cos^2 x + 3 \sin^2 x \right) \ dx &= \displaystyle \int_0^{{\pi}/{2}} \left( 2 \cos^2 x + 3 \right) \ dx \\ &= \displaystyle \int_0^{{\pi}/{2}} \left( \cos(2x) + 4 \right) \ dx \\ &= 2 \pi \end{aligned}$

The denominator can be solved as

$I_2 = \displaystyle \int_0^{{\pi}/{2}} \sin \theta \cos \theta \sqrt{25 \sin^2 \theta + 9 \cos^2 \theta} \ d\theta = \dfrac 12 \displaystyle \int_0^{{\pi}/{2}} \sin (2 \theta) \sqrt{16 \sin^2 \theta + 9 } \ d\theta$

Make substitution

$16 \sin^2 \theta + 9 = u \implies 16 \sin (2 \theta) \ d\theta = du$

So, our integral becomes

$\begin{aligned} \dfrac{1}{32} \displaystyle \int \sqrt{t} \ dt &= \dfrac{1}{48} t^{3/2} \\ &= \dfrac{1}{48} {(16 \sin^2 \theta + 9)} ^{3/2} \end{aligned}$

Putting required limits gives

$I_2 = \dfrac{98}{48} = \dfrac{49}{24}$

Thus, $\dfrac{I_1}{I_2} = \dfrac{2\pi}{{49}/{24}} = \boxed{\dfrac{48 \pi}{49}}$ .