Why this test statistics for these distributions For a random sample $X_1,\ldots,X_n$ from the Poisson distribution I get where the T-statistic comes from because for a Poisson $E(X) = \lambda$ and $V(X)=\lambda$ so it becomes:
$$T= \frac{\bar{X}-E(X)}{\sqrt{\frac{V(X)}{n}}} =\frac{\bar{X}-\lambda}{\sqrt{\frac{\lambda}{n}}},$$
where $\bar{X} = (X_1+\ldots,X_n)/n$.
But for the geometric distribution I have $E(X) = \frac{1}{p}$ and $V(X) = \frac{1}{p^2}$. And if I substitute:
$$T = \frac{\bar{X}-E(X)}{\sqrt{\frac{V(X)}{n}}} = \frac{\bar{X}-\frac{1}{p}}{\sqrt{\frac{1}{np^2}}} \neq \frac{\bar{X}-\frac{1}{p}}{\sqrt{\frac{1-p}{np^2}}}$$
The expression on the right is the real expression. Where does that $1-p$ in the numerator come from? How would I proceed with the binomial and exponential?

The variance of a geometric distribution is not $1/p^2$.  Suppose $X \sim \operatorname{Geometric}(p)$ with PMF $$\Pr[X = x] = p(1-p)^{x-1}, \quad x \in \{1, 2, \ldots \}.$$  (Note that the choice of support is irrelevant because the variance is independent of location.)  Then recall the formulas
$$g_1(z) = \sum_{k=1}^\infty kz^{k-1} = \frac{d}{dz}\left[\sum_{k=1}^\infty z^k\right] = \frac{d}{dz} \left[\frac{1}{1-z}\right] = \frac{1}{(1-z)^2}, \tag{1}$$
$$g_2(z) = \sum_{k=1}^\infty k(k-1)z^{k-1} = z \frac{dg_1}{dz} = \frac{2z}{(1-z)^3}. \tag{2}$$
Therefore,
$$\operatorname{E}[X] = \sum_{x=1}^\infty px(1-p)^{x-1} = p g_1(1-p) = \frac{p}{p^2} = \frac{1}{p}, \tag{3}$$ and
$$\operatorname{E}[X(X-1)] = pg_2(1-p) = \frac{2(1-p)}{p^2}. \tag{4}$$
It follows that
$$\begin{align}\operatorname{Var}[X] 
&= \operatorname{E}[X^2] - \operatorname{E}[X]^2 \\
&= \operatorname{E}[X(X-1) + X] - \operatorname{E}[X]^2 \\
&= \operatorname{E}[X(X-1)] + \operatorname{E}[X] - \operatorname{E}[X]^2 \\
&= \frac{2(1-p)}{p^2} + \frac{1}{p} - \frac{1}{p^2} \\
&= \frac{1-p}{p^2}. \tag{5}
\end{align}$$
