A definite integral is a limit of sums, and this works in both directions…

(a) $\displaystyle\sum_{k=1}^{n} \frac{n}{n^2 + 3k^2} = \frac1n \sum_{k=1}^{n} \frac{1}{1 + 3(k/n)^2} \to \int_0^1 \frac{dx}{1 + 3x^2} = \frac{1}{\sqrt3}\tan^{-1}\sqrt3 = \frac{\pi}{3\sqrt3} \approx 0.6046$, that is 0.60.

(b) $\displaystyle\sum_{k=1}^{2n} \frac{k}{n^2 + k^2} = \frac1n \sum_{k=1}^{2n} \frac{k/n}{1 + (k/n)^2} \to \int_0^2 \frac{x\,dx}{1 + x^2} = \frac12\ln 5 \approx 0.80$.

(c) $\dfrac{(3n)!}{n!} = (n + 1)(n + 2)\cdots(3n)$, a product of $2n$ factors, so the expression is $\displaystyle\prod_{k=1}^{2n}\left(1 + \frac kn\right)$. Its logarithm divided by $n$ is $\displaystyle\frac1n\sum_{k=1}^{2n}\ln\left(1 + \frac kn\right) \to \int_0^2 \ln(1 + x)\,dx = \Big[(1 + x)\ln(1 + x) - x\Big]_0^2 = 3\ln 3 - 2.$ The limit is $e^{3\ln 3 - 2} = \dfrac{27}{e^2} \approx 3.65$.

(d) Proof. For $k \le x \le k + 1$, since $f$ decreases, $f(k + 1) \le f(x) \le f(k)$, and the inequalities are strict for $k < x < k + 1$. Integrating over $[k, k + 1]$ gives $f(k + 1) < \displaystyle\int_k^{k+1} f(x)\,dx < f(k)$. Adding the right inequality for $k = 1, \dots, N$ gives $\displaystyle\int_1^{N+1} f < \sum_{k=1}^{N} f(k)$: the rectangles above the curve. Adding the left inequality for $k = 1, \dots, N - 1$ gives $\displaystyle\sum_{k=2}^{N} f(k) < \int_1^{N} f$; adding $f(1)$ to both sides gives the right half: the rectangles below the curve.

(e) With $f(x) = \dfrac{1}{\sqrt x}$ and $N = 400$: $2\left(\sqrt{401} - 1\right) < S < 1 + 2\left(\sqrt{400} - 1\right)$, that is $38.050 < S < 39$. So the integer part of $S$ is 38. (Adding the terms one by one gives $S \approx 38.565$.)