The first approach we take, though correct, is not the best one, and we later describe a better approach.
Suppose first that $x \ge \theta$. By symmetry, the probability that $X\le \theta$ is $\frac{1}{2}$. So if $x\ge \theta$, then
$$P(X\le x)= \frac{1}{2}+\int_{\theta}^x \frac{1}{2}e^{-(t-\theta)}\,dt.$$
The integration can be done by pulling out the $e^\theta$, but I prefer to make the substitution $u=t-\theta$. The integral becomes
$$\int_{u=0}^{x+\theta} \frac{1}{2}e^{-u}du,$$
which evaluates to
$$\frac{1}{2}(1-e^{-(x-\theta}).$$
Adding the $\frac{1}{2}$ for the probability that $X\le \theta$, we find that for $x\ge\theta$, $F_X(x)=1-\frac{1}{2}e^{-(x-\theta)}.$
For $x<\theta$, we need to find
$$\int_{-\infty}^x \frac{1}{2}e^{t-\theta}dt.$$
The integration is straightforward. We get that $F_X(x)=\frac{1}{2}e^{x-\theta}$ whenever $x <\theta$. We could go on the find the mean and variance by similar calculations, but will now change approach.
Another approach: The $\theta$ is a nuisance. Let's get rid of it. So let $Y=X-\theta$. Then $P(Y\le y)=P(X\le y-\theta)$. This is
$$\int_{-\infty}^{y-\theta} \frac{1}{2}e^{-|t-\theta|}dt.$$
Make the change of variable $w=t-\theta$. We find that our integral is
$$\int_{w=-\infty}^y \frac{1}{2}e^{-|w|}dw.$$
What this shows is the intuitively obvious fact that $Y$ has a distribution of the same family as the one for $X$, except that now the parameter is $0$. We could now repeat our integration work, with less risk of error. But that would be a waste of space, so instead we go on to find the expectation of $X$.
Since $X=Y+\theta$, we have $E(X)=E(Y)+\theta$. On the assumption that this expectation exists, by symmetry $E(Y)=0$, and therefore $E(X)=\theta$.
Next we deal with $\text{Var}(X)$. Since $X=Y+\theta$, the variance of $X$ is the same as the variance of $Y$. So we need to find
$$\int_{-\infty}^\infty \frac{1}{2}w^2e^{-|w|}dw.$$
By symmetry, this is twice the integral from $0$ to $\infty$, so we want
$$\int_0^\infty w^2e^{-w}dw.$$
Integration by parts (twice) handles this problem. To start, let $u=w^2$, and let $dv=e^{-w}dw$.
After a little while, you should find that the variance of $Y$, and hence of $X$, is $2$.
Remark: You can also find the mean and variance of $X$ by working directly with the original density function of $X$, and making an immediate substitution for $x-\theta$. But defining the new random variable $Y$ is in my view a more "probabilistic" approach.
Your distribution is a special case of the Laplace distribution, which in addition to a location parameter $\theta$, has a scale parameter $b$. The probability density function is
$$\frac{1}{2b}e^{-\frac{|x-\theta|}{b}}.$$
Continuing on Dilip's suggestion. PDF or a Normal RV is
$$
f(x)=\frac{1}{\sqrt{2 \pi} \sigma}e^{-\big( \frac{x- \mu}{\sigma \sqrt{2}} \big)^2}
$$
Using the values, you get $\mu=3, \sigma^2=8$, hence $X \sim N(3,8)$.
EDIT I probably misunderstood the OP's second question, as Dilip pointed out. Here's what I'd do. If $X \sim N(0,1)$, then:
$$
\mathbf{E}e^{-\frac{x}{2}}=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-\frac{x}{2}}e^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-\big(\frac{x^2}{2}+\frac{x}{2})}dx=\frac{e^{\frac{1}{8}}}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-\frac{1}{2} \big(x+ \frac{1}{2} \big)^2}dx = e^{\frac{1}{8}} \cdot 1\\
$$
The last step is since the integral is $\mathbf{P}(-\infty < X<\infty)$=1, where $X \sim N(-\frac{1}{2},1)$
Best Answer
You're finding the average value (not the average probability), and the value ranges from $-2$ to $3$, so $165/112$ looks like a possible average value (especially since the pdf is weighted toward the high numbers).