I was able to prove a) using the limit of $g$. I know that b) has something to do with Cauchy's formula, and I tried to use the RHS to get the LHS, but got stuck after expanding it.
I also tried using the hint, but I'm confused. Is $D[\bar{z}, r] = \bar {D}[z, r]$? What region would that represent? Also, even with the hint, I don't understand how that equals the RHS. Thank you for any help!
For (a), you can use Cauchy-Riemann equation. Let $f(x+iy)=u(x,y)+v(x,y)i$. Then, $g(x+iy)=u(x,-y)-v(x,-y)i=U(x,y)+V(x,y)i$. Since $f$ is holomorphic, $u$ and $v$ are differentiable with $u_x=v_y$ and $u_y=-v_x$, and so $U$ and $V$ are also differentiable. Now, $$U_x(x,y)=\frac{\partial}{\partial x}u(x,-y)=u_x(x,-y)=v_y(x,-y)=-\frac{\partial }{\partial y}v(x,-y)=V_y(x,y).$$ Finally, $$U_y(x,y)=\frac{\partial}{\partial y}u(x,-y)=-u_y(x,-y)=v_x(x,-y)=\frac{\partial }{\partial x}v(x,-y)=-V_x(x,y).$$ Thus, $U$ and $V$ also obey the Cauchy-Riemann conditions, and so $g$ is holomorphic.
Here is a proof of (b). With the new variable $\zeta=\frac1z$, we get $$\frac1{2\pi i}\int_{C[0,1]}\frac{\overline{f(z)}}{z-2}dz=\frac1{2\pi i}\int_{-C[0,1]}\frac{\overline{f\left(\frac{1}{\zeta}\right)}}{\frac{1}{\zeta}-2}\left(-\frac{d\zeta}{\zeta^2}\right)=\frac1{2\pi i}\int_{-C[0,1]}\frac{\overline{f\left(\frac{1}{\zeta}\right)}}{\frac{1}{\zeta}-2}\left(-\frac{d\zeta}{\zeta^2}\right)\,,$$ where $-C[0,1]$ denotes the clockwise (negatively oriented contour on $C[0,1]$. That is, $$\frac1{2\pi i}\int_{C[0,1]}\frac{\overline{f(z)}}{z-2}dz=\frac1{2\pi i}\int_{C[0,1]}\frac{\overline{f\left(\frac{1}{\zeta}\right)}}{\zeta(1-2\zeta)}d\zeta=\frac1{2\pi i}\int_{C[0,1]}\frac{\overline{f\left(\bar{\zeta}\right)}}{\zeta(1-2\zeta)}d\zeta\,,$$ as $\frac1\zeta=\bar\zeta$ for $\zeta\in C[0,1]$. That is, $$\frac1{2\pi i}\int_{C[0,1]}\frac{\overline{f(z)}}{z-2}dz=\frac1{2\pi i}\int_{C[0,1]}\frac{g(\zeta)}{\zeta(1-2\zeta)}d\zeta=\frac1{2\pi i}\int_{C[0,1]}\left(\frac1\zeta+\frac{2}{1-2\zeta}\right)g(\zeta)d\zeta\,.$$ Hence, $$\frac1{2\pi i}\int_{C[0,1]}\frac{\overline{f(z)}}{z-2}dz=\frac{1}{2\pi i}\int_{C[0,1]}\frac{g(\zeta)}{\zeta}d\zeta-\frac1{2\pi i}\int_{C[0,1]}\frac{g(\zeta)}{\zeta-\frac12}d\zeta=g(0)-g\left(\frac12\right).$$ Since $g(0)=\overline{f(\bar{0})}=\overline{f(0)}$ and $g\left(\frac12\right)=\overline{f\left(\overline{\left(\frac{1}{2}\right)}\right)}=\overline{f\left(\frac12\right)}$, we conclude that $$\frac1{2\pi i}\int_{C[0,1]}\frac{\overline{f(z)}}{z-2}dz=\overline{f(0)}-\overline{f\left(\frac12\right)}.$$ So, your problem is slightly wrong. There have to be complex conjugates at the end (you can verify that, without complex conjugates, the problem is false by checking that it fails for $f(z)=iz$).