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 $X \sim N(-\frac{1}{2},1)$
To solve these problems, you need to be aware of the geometry of the normal curve, and what the numbers in your standard normal table represent. Sometimes, they don't directly give you the number you want, and some manipulation is needed.
$1.$ I think the intention is to say let $Z$ be standard normal. Find $a$ so that $\Pr(|Z|\lt a)= .383$.
So the probability that $-a\lt Z\lt a$ is $.383$. Think of the standard normal curve. By symmetry, we want $-a\lt Z\lt 0=\Pr(0\lt Z\lt a=\frac{.383}{2}=.1915$.
So we want the area below $a$ to be $0.5+.1915=.6915$. If your table gives the probabilities that $Z\lt z$, look up $0.6915$ in the body of the table.
$2.$ The mean is $84$. You want $\Pr(|X-84|\gt 2.9$. This is the probability that $X\gt 84+2.9$ or less than $84-2.9$. So you want the probability that it differs from $84$ by more than $2.9$ in either direction, too low or too high. Find the individual probabilities and add. But by symmetry the two probabilities are the same. Can you find the probability that $X\gt 86.9$? If you do, double the result.
$3.$ The mean is $400$. We want the $k$ such that $\Pr(|X-400|\lt k)=0.975$. So In the two "tails", past $400+k$ and before $400-k$, we should have probability $0.025$. Thus each tail should have probability $0.0125$. What that means is that the total area below $400+k$ should be $1-0.0125=0.9875$. Perhaps you can take over from here.
$4.$ The probability that $35\lt X\lt 45$ is $\Pr(X\lt 45)-\Pr(X\lt 35)$. So the assertion that this probability is $0.65$ is just another way of saying that $\Pr(X\lt 45)=0.67$. Now to find $m$ and $s$, express the probability that $X\lt 35$ and the probability that $X\lt 45$ in terms of $m$ and $s$. It sounds as if you have done this for one of them. So it for the other and you will get $2$ linear equations in two unknowns. Solve.
Best Answer
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 $$\frac{1}{2b}e^{-\frac{|x-\theta|}{b}}.$$