(i) $$\int_{|z|=4}^{} \frac{e^{3z}}{z-i\pi} dz$$ Using Cauchy's integral formula I get $f(z)=e^{3z}, 2\pi if(i \pi)=-2\pi i$ Is this correct?
(ii) $$\int_{|z|=4}^{} \frac{e^{3z}}{(z+i\pi)^7} dz$$ For this I use the differentiation form of Cauchy's integral formula and get $\frac{2\pi i}{6!}f^{\text{6th derivative}}(-i \pi)=\frac{-729}{90}\pi i$. Is this correct? Not $100\text{%}$ about this one.
(iii)$$\int_{|z|=4}^{} \frac{e^{3z}}{2z-9i\pi} dz$$
For this I get $\frac{9\pi}{2}>\pi$ so integral is equal to $0$ using Cauchy's formula. Is this correct? Again not sure because I wasn't told about the point being outside the circle. Need clarification on this one.
Notice,
1) $z=i\pi$ is a first order pole & $f(z)=e^{3z}$ hence $$\int_{|z|=4}\frac{e^{3z}}{z-i\pi}\ dz=2\pi i \left(e^{3z}\right)_{z=i\pi}=2\pi i e^{3i\pi}=2\pi i(\cos 3\pi+i\sin 3\pi)=-2\pi i$$ your answer (1) is correct
2) $z=i\pi$ is a seventh order pole & $f(z)=e^{3z}$ hence $$\int_{|z|=4}\frac{e^{3z}}{z-i\pi}\ dz=\frac{2\pi i}{6!} \left(\frac{d^6}{dz^6}(e^{3z})\right)_{z=i\pi}=\frac{2\pi i}{6!}(3^6) e^{3i\pi}=\frac{81\pi i}{40}(\cos 3\pi+i\sin 3\pi)=-\frac{81\pi i}{40}$$ your answer (2) is correct
3) $z=\frac{9}{2}i\pi $ is a first order pole lying outside the circle $|z|=4$ hence
$$\int_{|z|=4}\frac{e^{3z}}{2z-9i\pi}\ dz=0$$ your answer (3) is correct