a) Clearly, $K = (a_{2n-1} + a_{2n})/a_n$ is a positive rational number. In fact, $K$ must be integral. To prove this, write $K = p/q$ in lowest terms to deduce that the $a_n$ are all divisible by $q$. Divide them all by $q$ to obtain a new sequence whose corresponding ratios are again $K$. Repetition of the process to the new sequence and its successors shows that the $a_n$ are all divisible by arbitrarily large powers of $q$, so $q=1$ and $K$ is indeed integral.
Since the $a_n$ form a strictly increasing sequence, it follows that $K > 2$, and since the latter is integral, it is at least $3$.
To rule out the case $K = 3$, we consider the positive integers $b_n = a_{n+1} - a_n$, show that for every index $m$ there exists an index $n > m$ such that $b_n < b_m$ and reach thereby a contradiction. Indeed, if $K = 3$, then $3b_n = b_{2n-1} + 2b_{2n} + b_{2n+1}$, so at least one of the three $b$'s in the right-hand member must be less than $b_n$. Consequently, $K \ge 4$.

b) If $N = 4$, let $a_n = 2n - 1$; in this case, the verifications are obvious. If $N \ge 5$, set $a_1 = 1$ and let $a_{2n-1} = \lfloor (N a_n - 1)/2 \rfloor$ and $a_{2n} = \lfloor N a_n/2 \rfloor + 1$. This sequence satisfies the required ratio condition, $a_{2n-1}$ is obviously less than $a_{2n}$, and it is sufficient to prove that $a_{2n} < a_{2n+1}$. This can be done by noticing that $a_2 < a_3$, and showing that if $a_n < a_{n+1}$, then $a_{2n} < a_{2n+1}$. Indeed, $a_{2n+1} - a_{2n} \ge (N a_{n+1} - 2)/2 - (N a_n/2 + 1) = N(a_{n+1} - a_n)/2 - 2 \ge N/2 - 2 \ge 1/2$.