You should know that using the residue theorem would be easier, but if we are to restrict ourselves to Cauchy's integral formula, then here's one way of attacking it:
First, note that the integrand is a quotient of two entire functions. As such, the integrand is analytic everywhere except the points at which the denominator is zero. Since $z^2+4$ can be factored as $(z-2i)(z+2i)$, then the only points at which the integrand is not analytic are $\pm 2i$. Unfortunately, both of these points are inside the circle $|z| = 3$, so in order to apply Cauchy's integral formula, we will have to be clever.
Let $C$ be the circle $|z| = 3$ oriented counterclockwise. Let $C_1$ be the upper half of the circle $|z| = 3$ together with the line segment $[-3, 3]$ oriented counterclockwise. Let $C_2$ be the lower half together with the line segment $[-3, 3]$ oriented counterclockwise. Notice we have:
$$\int_C \frac{e^{zt}}{z^2+4} \ dz = \int_{C_1} \frac{e^{zt}}{z^2+4} \ dz + \int_{C_2} \frac{e^{zt}}{z^2+4} \ dz$$
Now we can attack the two integrals on the right hand side separately using Cauchy's integral formula. For the first, e.g., you can let $\displaystyle f(z) = \frac{e^{zt}}{z+2i}$, which is analytic everywhere inside $C_1$, and your integrand becomes $\displaystyle \frac{f(z)}{z-2i}$.
Alternative method:
Using partial fraction decomposition, we have $\displaystyle \frac{1}{z^2+4} = \frac{i}{4(z+2i)} - \frac{i}{4(z-2i)}$. Hence:
$$\int_C \frac{1}{z^2+4} \ dz = \int_C \frac{i}{4(z+2i)} \ dz - \int_C \frac{i}{4(z-2i)} \ dz$$
And then one can apply Cauchy's integral formula on the two separate pieces without having to split the contour.
Let us first consider Part (ii).
As the only singularity of the function $f(z)$ is at $z = 3$, any integral over a closed curve that does not contain the point $z = 3$ is just $0$, as there are no singularities contained within the curve.
Therefore, $\oint_C f(z)dz = 0$ for any closed curve $C$ that doesn't contain the point $z = 3$.
Now we consider Part (i), where $z = 3$ is contained within the curve. Here, one must evaluate the Residue of $f(z)$ the point $z = 3$.
The Residue is easily evaluated, as we only have to deal with a simple pole. It is given by: $$\text{Res}_3 f(z) = 2\pi i\lim\limits_{z \rightarrow 3}{(z - 3)f(z)} = 2\pi i$$
Thus, in this case, $\oint_C f(z)dz = 2\pi i$, if I haven't made any mistakes (no guarantees there, I'm pretty new to Residue Calculus myself).
Best Answer
Use the residue theorem. Since $\cfrac{z^3}{(z-i)^n}$ has a pole of order $n$ at $z=i$ and analytic everywhere other than $z=i$ in the domain $|z-i|<1$, by residue theorem, we have $$\int_{\gamma}\frac{z^3}{(z-i)^n}dz=2\pi ig(i),\text{where }g(z)=\frac{1}{(n-1)!}(z^3)^{(n-1)}$$
The residue theorem is obtained from Cauchy Integral formula.
By Cauchy Integral formula, we have $$2\pi if(z)=\int_C\frac{f(\zeta)}{\zeta-z}d\zeta$$ Differentiate both sides with respect to $z$ $n-1$ times, we get $$2\pi if^{(n-1)}(z)=(n-1)!\int_C\frac{f(\zeta)}{(\zeta-z)^n}d\zeta$$