Applying Itô's formula to solve this problem is kind of overkill. It is well-known that for any Gaussian random variable $X$ with mean $0$ and variance $\sigma^2$ it holds that
$$\mathbb{E}e^{\xi X} = \exp \left( \frac{1}{2} \sigma^2 \xi^2 \right) \qquad \text{for all $\xi \in \mathbb{R}$.}$$
Since $B_T \sim N(0,T)$ this implies
$$\mathbb{E}e^{B_T} = e^{T/2} \qquad \text{and} \qquad \mathbb{E}e^{2 B_T} = e^{2T}.$$
Using that $\text{var}(X) = \mathbb{E}(X^2)-(\mathbb{E}(X))^2$ we get
$$\text{var}(e^{B_T}) = \mathbb{E}(e^{2B_T}) - (\mathbb{E}e^{B_T})^2 = e^{2T}-e^{T}.$$
Regarding your attempt: Using that
$$e^{B_T} = e^{T/2} + e^{T/2} + e^{T/2} \int_0^T e^{B_s-s/2} \, dB_s$$
we find by Itô's isometry
$$\begin{align*} \text{var}(e^{B_T}) &= \mathbb{E} \big[ (e^{B_T} - e^{T/2})^2 \big] \\ &= e^T \mathbb{E} \left| \int_0^T e^{B_s-s/2} \, dB_s \right|^2 \\ &= e^T \mathbb{E} \left( \int_0^T e^{2B_s-s} \, ds \right) \\ &= \int_0^T e^{-s} \mathbb{E}e^{2B_s} \, ds. \end{align*}$$
In order to calculate the remain integral you have to calculate $\mathbb{E}e^{2B_s}$. Since we have already seen in the first part of my answer that we can easily calculate the variance of $e^{B_t}$ if we know $\mathbb{E}e^{2B_t}$, this approach doesn't make any sense, from my point of view.
For the first question, I will outline a proof of the fact that if $W_s$ is a real-valued Brownian motion then $C = \int_0^\infty e^{2W_s} ds = \infty$. This is sufficient since $\int_0^t e^{2W_s}ds$ is increasing in $t$.
The most simple first guess for how to do this is to bound $C \geq \int_0^\infty 1_{\{W_s \geq 0\}} ds$ and try to show that there almost surely exists an infinite sequence of times $t_n > 1$ such that $W(s) \geq 0$ for all $s \in [t_n - 1, t_n]$ (note that $t_n$ is a random variable here).
One can check that $T = \inf\{t > 1: W_u \geq 0 \text{ for all } u \in [t-1, t]\}$ is a stopping time for the Brownian filtration and so to see the existence of the sequence $t_n$ it suffices to show that $\mathbb{P}(T < \infty) = 1$ and apply the usual recursive argument using the strong Markov property.
Let $H_n = \inf\{t > 0: W_t = n\}$. Then, by the strong Markov property and symmetry, $$\mathbb{P}(W_t < 0 \text{ for some } t \in [H_n, H_n + 1]) = \mathbb{P}( H_n \leq 1).$$
By the reflection principle, $\mathbb{P}(H_n \leq 1) = 2 \mathbb{P}(W_1 \geq n) \to 0$ as $n \to \infty$ since $W_1$ is a standard Gaussian.
In particular, we can deduce that
$$\mathbb{P}(\forall n, W_t < 0 \text{ for some } t \in [H_n, H_n + 1]) = 0$$
which in turn implies that $$\mathbb{P}(T < \infty) \geq \mathbb{P}( \exists n, W_t \geq 0 \text{ for all } t \in [H_n,H_n + 1]) = 1.$$
For your second question, if $f(B_s)$ is a time-change of the Brownian motion $\tilde{B}_s$ then the events $A = \{\exists s: f(B_s) = 0\}$ and $B = \{\exists u: \tilde{B}_u = 0\}$ are equal. Since $f(B_s) = e^{B_s}$, you know that $\mathbb{P}(A) = 0$ which implies that $\mathbb{P}(B) = 0$ which is the desired result (for the Brownian motion $\tilde{B}$).
Best Answer
Consider the processes $$X_t=\int ^t_0Y_s dB_s\qquad Y_t=(B_t+1)^2$$ By repeated applications of Itô isometry, one gets:
Finally, if one can compute $E(Y_t)$ and $E(B_tY_t)$, the proof is complete. Can you?