1. We need to solve the linear differential equation

$\frac{dy}{dx}-y=1+4\sin x,$ with the condition

$y(\pi)=1.$

We then have to find

$y\left(\frac{\pi}{2}\right)+10.$

2. Write the equation in standard linear form:

$\frac{dy}{dx}+P(x)y=Q(x),$ where here

$P(x)=-1, \qquad Q(x)=1+4\sin x.$

So the integrating factor is

$\mathrm{I.F.}=e^{\int -1\,dx}=e^{-x}.$

3. Multiply the differential equation by $e^{-x}$:

$e^{-x}\frac{dy}{dx}-e^{-x}y=(1+4\sin x)e^{-x}.$

The left side becomes

$\frac{d}{dx}(ye^{-x})=(1+4\sin x)e^{-x}.$

Thus,

$ye^{-x}=\int (1+4\sin x)e^{-x}\,dx+C.$

4. Compute the integral:

$\int (1+4\sin x)e^{-x}\,dx=\int e^{-x}\,dx+4\int e^{-x}\sin x\,dx.$

First,

$\int e^{-x}\,dx=-e^{-x}.$

Now use the standard result

$\int e^{ax}\sin bx\,dx=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx).$

Here $a=-1$, $b=1$, so

$\int e^{-x}\sin x\,dx=\frac{e^{-x}}{2}(-\sin x-\cos x).$

Therefore,

$4\int e^{-x}\sin x\,dx=4\cdot \frac{e^{-x}}{2}(-\sin x-\cos x)= -2e^{-x}(\sin x+\cos x).$

So,

$\int (1+4\sin x)e^{-x}\,dx=-e^{-x}-2e^{-x}(\sin x+\cos x).$

Hence,

$ye^{-x}=-e^{-x}-2e^{-x}(\sin x+\cos x)+C.$

Multiply by $e^x$:

$y=-1-2(\sin x+\cos x)+Ce^x.$

So,

$y=Ce^x-1-2\sin x-2\cos x.$

5. Use the condition $y(\pi)=1$:

Since

$\sin \pi=0, \qquad \cos \pi=-1,$ we get

$1=Ce^{\pi}-1-2(0)-2(-1)=Ce^{\pi}-1+2=Ce^{\pi}+1.$

Thus,

$Ce^{\pi}=0 \implies C=0.$

So the solution is

$y=-1-2\sin x-2\cos x.$

6. Now evaluate at $x=\frac{\pi}{2}$:

$\sin \frac{\pi}{2}=1, \qquad \cos \frac{\pi}{2}=0.$

Hence,

$y\left(\frac{\pi}{2}\right)=-1-2(1)-2(0)=-3.$

Therefore,

$y\left(\frac{\pi}{2}\right)+10=-3+10=7.$

7. Final answer:

$\boxed{7}$