Question 1: Let $H$ be the event "buys high-tech" and $M$ the event "buys every month."
(i) We have been told that $P(H|M)=0.3$. If $H$ and $M$ were independent, we would have $P(H|M)=P(H)$. But we have been told that $P(H)=0.2$. So $H$ and $M$ are not independent.
Much more informally, the proportion of high-tech shoppers among monthly shoppers is $0.3$, substantially more than the proportion of high-tech shoppers in the general population. If we know that someone is a monthly shopper, our estimate that she is a high-tech shopper is different (and bigger) that if we do not know about the monthly shopping habit.
(ii) We want $P(M|H)$, the probability of $M$ given $H$. But $P(M|H)P(H)=P(M\cap H)=P(H|M)P(M).$ We know $P(H|M)$, and $P(M)$, and $P(H)$, so we can compute $P(M|H)$. The answer turns out to be $0.27$.
We can also use the fact that $P(H\cap M)=P(H|M)P(M)$ to find that $P(H\cap M)=(0.3)(0.6)=0.18$. But $P(H)P(M)=(0.2)(0.6)=0.12$. So $P(H\cap M)\ne P(H)P(M)$, which is another (and in this case more complicated) way of seeing that $H$ and $M$ are not independent.
Question 2: The procedure for (i) is right.
For (ii), if $Y$ is the number of smokers in a sample of $400$, then $Y$ has binomial distribution, mean $(400)(0.25)$ and variance $(400)(0.25)(0.75)=75$. The probability that $Y \le 112$ is, approximately, the probability that a normal with mean $100$ and variance $75$ is $\le 112.5$. (We have made the continuity correction. If you do not, and use $112$ instead, the approximation is likely to be less good.)
So our probability is approximately the same as the probability that $Z\le \frac{112.5-100}{\sqrt{75}}$, where $Z$ is standard normal.
Added Remarks: I do not think that "less than $113$" can be interpreted as meaning $113$ or fewer, which is where your $1.5$ comes from. By the way, the inequality should go the other way, we want (with your interpretation) $P(Z\lt 1.5)$.
The probability that $Y \le 112$ is (Wolfram Alpha) approximately $0.924184$. With the continuity correction, the normal approximation gives probability roughly $0.925543$. Not bad. The probability that the normal is $\le 112$ (so no continuity correction) is about $0.917072$, respectable, but not nearly as accurate.
Note that with the availability of good computing tools, we can evaluate binomial probabilities directly, so the normal approximation to the binomial is of diminishing practical importance.