Confidence Interval Probabilty 
1) We have discovered a new method to produce a liquid containing
  Zinc. We measure the concentration μ four times independently (from
  the same solution) and find the values
0.3, 0.33, 0.3, 0.27
a) Assuming that the standard deviation of the measurement error is
  0.02 (when we make one measurement) estimate μ and give a 90%-confidence interval.
Answer:
We assume the measurement errors to be normal with zero expectation.
  The estimate for μ is
$μ_X = (X_1 + X_2 + X_3 + X_4)/4 = (0.3 + 0.33 + 0.3 + 0.27)/4 = 0.3.$
The same procedure as usual leads to a confidence interval
$[0.3 - b_\text{standard deviation}/\sqrt{n},\, 0.3 + b_\text{standard deviation}/\sqrt{n}]$
where the standard deviation equals $0.02$ and $\sqrt{n} = 2$,
  whilst $b$ is defined by the equation
$P(−b \leq N(0, 1)\leq b) = 90\%.$
The last equation above is equivalent to $\lambda(b) = 0.95$ which we can solve with the help of a table and find $b = 1.645$. Hence the $90\%$ confidence interval is equal
  to $[0.3 − 0.0164, 0.3 + 0.0164] = [0.2836, 0.3164]$. The true
  concentration is thus between $0.2836$ and $0.3164$ with at least $95\%$
  probability.

My question is how were they able to get $95\%$ probability and $b$?
 A: $0.95$ comes from the fact that $\lambda(b) = 1 - \lambda(-b)$ and that you want $b$ to satisfy $\lambda(b) - \lambda(-b) = 0.9$. The number $b$ is found by looking up in the table of values of $\lambda$.
(I assume that $\lambda$ is the cumulative distribution function of the normal distribution with zero mean and variance $1$.)
A: You want to find $b$ such that $P(−b \leq N(0, 1)\leq b) = 0.9$.  From the symmetry of the normal density function, this means that 
$$P(b < N(0,1) < \infty) = P(-\infty < N(0,1) < b) = 0.05$$
since $10\%$ of the area under the normal density function is outside the
interval $[-b,b]$.  You have a table of values of the cumulative standard
normal distribution function, which tells you the value of
$$P(-\infty < N(0,1) \leq b) = P(-\infty < N(0,1) < b) + P(−b \leq N(0, 1)\leq b)
= 0.05 + 0.9 = 0.95$$
So you look in the table and find that $b = 1.645$ gives a value of $0.9495$,
just a little less than the desired $0.9500$,
and so the exact value of $b$ is a tad more that $1.645$.
A: I'm not fully certain what the question is, but for I'll guess that it's something like this: The probability that a normally distributed random variable is less than its expected value plus $1.645$ standard deviations is about $95\%$, so how is that related to the $90\%$ figure mentioned after that?
$$
\Pr(Z<1.645) = 0.95.
$$
So
$$
\Pr(Z>1.645) = 0.05.
$$
By symmetry, then,
$$
\Pr(Z<-1.645) = 0.05.
$$
So you have $5\%$ of the probability in the upper tail and $5\%$ in the lower tail.  That makes $10\%$.  So $90\%$ remains in the middle.
