First hint. Use the inequality: $e^{x - x^2} \leq 1+x \leq e^{x}$ valid for $|x| \leq 1/4$.
Full Solution. As a first step, I'll use the hint to get the inequality @George suggests: $ 1+\frac{a}{n}+\frac{bk}{n^2} \leq \exp(\frac{a}{n}+\frac{bk}{n^2}), $ and $ 1+\frac{a}{n}+\frac{bk}{n^2} \geq \exp(\frac{a}{n}+\frac{bk}{n^2} - (\frac{a}{n}+\frac{bk}{n^2})^2) \geq \exp(\frac{a}{n}+\frac{bk}{n^2} - \frac{(a+b)^2}{n^2}). $
Now we will use this to get the limit. We have the following upper bound on the product: $ \prod_{k=1}^n (1+\frac{a}{n}+\frac{bk}{n^2}) \leq \prod_{k=1}^{n} \exp(\frac{a}{n}+\frac{bk}{n^2}) = \exp(\frac{a}{n}n + \frac{b}{n^2} \frac{n(n+1)}{2}) = \exp(a+\frac{b}{2}+\frac{b}{2n}). $ We can also establish a similar lower bound: $ \prod_{k=1}^n (1+\frac{a}{n}+\frac{bk}{n^2}) \geq \prod_{k=1}^{n} \exp(\frac{a}{n}+\frac{bk}{n^2} - \frac{(a+b)^2}{n^2}) = \exp(a + \frac{b}{2} + \frac{b}{2n} - \frac{(a+b)^2}{n}). $ Note that both the upper and lower bounds approach the limit $\exp(a+\frac{b}{2})$ as $n \to \infty$. Therefore, by the squeeze theorem, the given product also has the same limit.