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.
If we substitute $\mu_1=0$ and $\mu_2=-1$ in the exponent of $e $ of the joint probability density formula, we get
$$\displaystyle -\frac 1{2(1-\rho^2)}\left[\left(\frac{x}{\sigma_1}\right)^2
-2\rho \left(\frac{x}{\sigma_1}\right) \left(\frac{y+1}{\sigma_2}\right) +\left(\frac{y+1}{\sigma_2}\right)^2\right]$$
Expanding it, we have
$$-\frac 1{2(1-\rho^2)} \left[\frac{1}{\sigma_1^2}x^2 +\frac{1}{\sigma_2^2} y^2
- \frac{ 2\rho }{\sigma_1 \sigma_2} xy - \frac{ 2\rho }{\sigma_1 \sigma_2} x + \frac{2y}{\sigma_2^2} + \frac{1}{\sigma_2^2} \right]$$
which can be written, if we move a given factor $k^2$ into the brackets, as
$$-\frac {1}{2k^2(1-\rho^2)} \left[\frac{k^2}{\sigma_1^2}x^2 +\frac{k^2}{\sigma_2^2} y^2
- \frac{ 2k^2 \rho }{\sigma_1 \sigma_2} xy - \frac{ 2 k^2\rho }{\sigma_1 \sigma_2} x + \frac{2k^2y}{\sigma_2^2} + \frac{k^2}{\sigma_2^2} \right]$$
Comparing this expression with
$$-\frac {1}{54} \left[x^2 +4 y^2+2 xy +2 x + 8y + 4 \right]$$
which is provided in the OP we directly get $\sigma_1=k \,\,$, $\sigma_2=k/2\,\,$ (note that both variances are by definition positive), and $\rho=-1/2\,\,$. So we have
$$-\frac {1}{2k^2 \left(1-\left (\frac {1}{2} \right)^2 \right)} =-\frac {1}{54} $$
and then $k=\pm 6\,\,$. This leads to $\sigma_1^2=36\,\,$ and $\sigma_2^2=9\,\,$.
Best Answer
If $X$ is a normally distributed random variable, then so is $g(X) = mX+b$. But the proposition as you stated it is wrong.
A proof can go like this (first assume $m>0$, then mutatis mutandis): $$ \begin{align} F_{g(X)}(y) & = \Pr(g(X) \le y) = \Pr(mX+b \le y) = \Pr\left(X \le \frac{y-b}{m}\right) \\[12pt] & = \int_{-\infty}^{(y-b)/m} \frac{1}{\sqrt{2\pi}}\cdot\exp\left(\frac{-(x-\mu)^2}{2\sigma^2}\right)\,\frac{dx}{\sigma} ; \end{align} $$ therefore $$ f_{g(X)}(y) = \frac{d}{dy} \int\cdots\cdots\text{(ditto)}\cdots\cdots\quad = \frac{1}{\sqrt{2\pi}} \exp \left( \frac{-\left(\frac{y-b}{m}-\mu\right)^2}{2\sigma^2}\right)\frac{1}{\sigma}\cdot\frac{d}{dy}\,\frac{y-b}{m}, $$ and then simplify.