Correlation and Covariance The book I'm reading gives this as an example for lognormal variables.
Starting at some fixed time, let $S(n)$ denote the price of a security at the end of $n$ additional weeks, $n \ge 1$. A popular model for the evolution of these prices assumes that $S(n)/S(n-1)$ for $n \ge 1$ are independently and identically distributed lognormal random variables. 
Assuming this model, with lognormal parameters μ = .0165 and σ = .0730, what is the probability that:
(a.) The price of the security increases over each of the next two weeks
Solution part a
(b.) The price at the end of two weeks is higher than today?
Solution part b
What I'm confused about is:
Part A: Why is it not $\frac{P(2)}{P(1)} \cdot \frac{P(1)}{P(0)}$?
How did it go from $\log(S(1)/S(0))$ to $-0.0165/0.0730$?
Part B: Again, how did they convert from $\log(S)$ to $Z > -0.0330/0.0730 \sqrt{2}$
(step 2 -> step 3)
I'm just starting to learn this material. 
 A: I agree that the solution provided contains an oversight, if the desired probability is that the price of the security increases over two successive weeks.  This would require $$\frac{S(2)}{S(1)} > 1, \quad \text{and} \quad \frac{S(1)}{S(0)} > 1.$$  But this is easily remedied:  because the question stipulates that these ratios are IID lognormal variables, the probability that both ratios are greater than 1 is simply the product of their individual probabilities; i.e., $$\Pr \left[ \frac{S(2)}{S(1)} > 1 \cap \frac{S(1)}{S(0)} > 1\right] = \Pr\left[\frac{S(1)}{S(0)} > 1\right]^2.$$  And because this last probability is equivalent to $$\Pr\left[\log\frac{S(1)}{S(0)} > 0\right],$$ and the logarithm of a lognormal variable is normally distributed with the mean and standard deviation equal to what was given in the question, the correct answer is simply the square of what was given, or approximately $0.347404$.
The solution to the second part only requires that $S(2)/S(0) > 1$; it doesn't require $S(1) > S(0)$, or $S(2) > S(1)$.  For example, the price at week 1 could be less than at the start, but it increases more in week 2 such that the net change over two weeks is positive; or, it could increase by a lot in week 1, and decrease by a little in week 2, yet the net change could still be positive.  So perhaps the best way to do this is to note that $$\frac{S(2)}{S(0)} = \frac{S(2)}{S(1)} \cdot \frac{S(1)}{S(0)},$$ and the RHS is the product of two lognormal variables; thus the logarithm of $S(2)/S(0)$ is the sum of two normal variables, both of which are independent and identically distributed with the given mean and standard deviation.  That is to say, $$\Pr\left[\frac{S(2)}{S(0)} > 1\right] = \Pr[X_1 + X_2 > 0],$$ where $$X_i = \log \frac{S(i)}{S(i-1)} \sim \operatorname{Normal}(\mu,\sigma^2).$$  And since $X_1$, $X_2$ are IID normal, their sum is also normal with mean $2\mu$ and standard deviation $\sigma\sqrt{2}$.  (Remember, the variance of a sum of independent random variables equals the sum of the variances of each variable.)  This is where the $\sqrt{2}$ factor comes in.
