Answer: \(a=\frac12\ln101\); \(y=3x-\frac25\sin5x+\pi\); \(\int_{\pi/2}^{\pi}y\,dx=\frac{13\pi^2}{8}-\frac{2}{25}\).

First integrate:

\(\int e^{2x}\,dx=\frac12e^{2x}.\)

Using the limits,

\(\int_0^a e^{2x}\,dx =\left[\frac12e^{2x}\right]_0^a =\frac12e^{2a}-\frac12.\)

This is equal to 50, so

\(\frac12e^{2a}-\frac12=50.\)

Therefore

\(e^{2a}=101,\)

and hence

\(\boxed{a=\frac12\ln101}.\)

For the curve, integrate

\(\frac{dy}{dx}=3-2\cos5x.\)

This gives

\(y=3x-\frac25\sin5x+C.\)

The curve passes through \(\left(\frac{\pi}{5},\frac{8\pi}{5}\right)\). Substitute this point:

\(\frac{8\pi}{5}=3\cdot\frac{\pi}{5}-\frac25\sin\pi+C.\)

Since \(\sin\pi=0\),

\(C=\pi.\)

Therefore

\(\boxed{y=3x-\frac25\sin5x+\pi}.\)

Now integrate \(y\):

\(\int y\,dx =\int\left(3x-\frac25\sin5x+\pi\right)\,dx.\)

So

\(\int y\,dx=\frac{3x^2}{2}+\frac{2}{25}\cos5x+\pi x+C.\)

Hence

\(\int_{\pi/2}^{\pi}y\,dx =\left[\frac{3x^2}{2}+\frac{2}{25}\cos5x+\pi x\right]_{\pi/2}^{\pi}.\)

At \(x=\pi\), this is

\(\frac{3\pi^2}{2}+\frac{2}{25}\cos5\pi+\pi^2 =\frac{5\pi^2}{2}-\frac{2}{25}.\)

At \(x=\frac{\pi}{2}\), this is

\(\frac{3\pi^2}{8}+\frac{2}{25}\cos\frac{5\pi}{2}+\frac{\pi^2}{2} =\frac{7\pi^2}{8}.\)

Therefore

\(\int_{\pi/2}^{\pi}y\,dx =\frac{5\pi^2}{2}-\frac{2}{25}-\frac{7\pi^2}{8} =\boxed{\frac{13\pi^2}{8}-\frac{2}{25}}.\)