Question is: Calculate the mechanical work of the force $$ \vec{F}=(y^2+z^2)\vec{i}+(x+y)\vec{k}$$
which acts on a point moving on the curve $$\gamma: \begin{cases}x^2+y^2+z^2=4x\\ x^2+y^2=2x\\z≥0\end{cases}$$
Answer:
$$\gamma: \begin{cases}x(t)=2cos^2t\\ y(t)=2cost\cdot sint\\z(t)=2cost\end{cases}$$ where $$t \in [-\frac{\pi }{2},\frac{\pi }{2}]$$
then we have $$\gamma: \begin{cases}x'(t)=-2sin2t\\ y'(t)=2cos2t\\z'(t)=-2sint\end{cases}$$ where $$t \in [-\frac{\pi }{2},\frac{\pi }{2}]$$
Calculate the integral now: $$\int _{\gamma }\:\left(y^2+z^2\right)dx+\left(x+y\right)dz$$ =$$\int _{-\frac{\pi }{2}}^{\frac{\pi \:}{2}}\:\left[\left(\left(2costsint\right)^2+\left(2cost\right)^2\right)\left(-2sint\right)+\left(2cos^2t+2cost\cdot sint\right)\left(-2sint\right)\right]dt$$ = $$-\frac{8}{3}$$
There is one mistake in your working though it does not change the answer, as that part of the integral is anyway zero. Please see error highlighted in red.
$\displaystyle \int _{-\pi/2}^{\pi/2} \left[(4 \sin^2 t \cos^2 t + 4 \cos^2 t) \left(\color {red} {-2 \sin t} \right) + \left(2 \cos^2 t + 2 \sin t \cos t \right)\left(-2 \sin t\right)\right]dt$
It should instead be,
$\displaystyle \int _{-\pi/2}^{\pi/2} \left[(4 \sin^2 t \cos^2 t + 4 \cos^2 t) \left(\color {blue} {- 4 \sin t \cos t} \right)+\left(2 \cos^2 t + 2 \sin t \cos t \right)\left(-2 \sin t\right)\right]dt$
As there is symmetry about xz plane and $\sin t$ and $\sin ^3 t$ are odd functions, the integral reduces to,
$\displaystyle \int _{-\pi/2}^{\pi/2} - 4 \sin^2 t \cos t \ dt = - \frac{8}{3}$