The second problem (d), where the mean is equal to the variance is discussed on pp. 53 of Asymptotic Theory of Statistics and Probability by Anirban DasGupta (2008). The $\mathcal{N}(\theta, \theta)$ distribution, the normal distribution with an equal mean and variance can be seen as a continuous analog of the Poisson distribution.
I will try to outline a path to the solutions.
The log-likelihood function of a $\mathcal{N}(\mu, \sigma^{2})$ is given by:
$$
\ell(\mu, \sigma^2)=-\frac{n}{2}\log(2\pi)-\frac{n}{2}\log(\sigma^2)-\frac{1}{2\sigma^{2}}\sum_{i=1}^{n}(x_{i}-\mu)^{2}.
$$
Setting $\mu=\sigma^{2}=\theta$ yields
$$
\ell(\theta)=-\frac{n}{2}\log(2\pi)-\frac{n}{2}\log(\theta)-\frac{1}{2\theta}\sum_{i=1}^{n}(x_{i}-\theta)^{2}.
$$
Expanding the term under the sum leads to
$$
\begin{align}
\ell(\theta) &=-\frac{n}{2}\log(2\pi)-\frac{n}{2}\log(\theta)-\frac{1}{2\theta}\left(\sum_{i=1}^{n}x_{i}^{2}-2\theta\sum_{i=1}^{n}x_{i}+n\theta^{2}\right) \\
&=-\frac{n}{2}\log(2\pi)-\frac{n}{2}\log(\theta)-\frac{s}{2\theta}+t-\frac{n\theta}{2} \\
\end{align}
$$
where $s=\sum_{i=1}^{n}x_{i}^{2}$ and $t=\sum_{i=1}^{n}x_{i}$. Taking the first derivative wrt $\theta$ gives
$$
S(\theta)=\frac{d}{d\theta}\ell(\theta)=\frac{s}{2\theta^{2}}-\frac{n}{2\theta}-\frac{n}{2}.
$$
So $s$ is the minimal sufficient statistic. The maximum likelihood estimator $\hat{\theta}$ can be found by setting $S(\theta)=0$ and solving for $\theta$.
Applying the same procedure to $\mathcal{N}(\mu=\theta, \sigma^{2}=\theta^{2})$, the log-likelihood function is
$$
\ell(\theta)=-\frac{n}{2}\log(2\pi)-\frac{n}{2}\log(\theta^{2})-\frac{1}{2\theta^{2}}\sum_{i=1}^{n}(x_{i}-\theta)^{2}.
$$
This leads to the following score function (again, with $s=\sum_{i=1}^{n}x_{i}^{2}$ and $t=\sum_{i=1}^{n}x_{i}$):
$$
S(\theta)=\frac{s}{\theta^{3}}-\frac{t}{\theta^2}-\frac{n}{\theta}.
$$
Fisher information
The Fisher information is defined as the negative second derivative of the log-likelihood function:
$$
I(\theta)=-\frac{d^{2}\,\ell(\theta)}{d\,\theta^{2}}=-\frac{d\,S(\theta)}{d\,\theta}.
$$
The observed Fisher information is $I(\hat{\theta})$, the Fisher information evaluated at the maximum likelihood estimate.
For the second question (d), we have:
$$
I(\theta)=-\frac{d}{d\,\theta}\left(\frac{s}{2\theta^{2}}-\frac{n}{2\theta}-\frac{n}{2} \right) = \frac{s}{\theta^{3}}-\frac{n}{2\theta^{2}}.
$$
And for the first question (c), we have:
$$
I(\theta)=-\frac{d}{d\,\theta}\left(\frac{s}{\theta^{3}}-\frac{t}{\theta^2}-\frac{n}{\theta}\right) = \frac{3s}{\theta^{4}}-\frac{2t}{\theta^{3}}-\frac{n}{\theta^{2}}.
$$
To get the observed Fisher information, plug in the maximum likelihood estimates.
Gamma distribution
It looks right to me but you don't need the sums in the expressions of the Fisher information.
