(a): To use induction, you need to prove the equation for the base case (in this case $n=1$), then prove that given the equation is true for $n=m$, it is true for $n=m+1$ as well. In other words, you show that (eqn. is true for $n=m$) implies (eqn. is true for $n=m+1$). Then, since the equation is true for $n=1$, it must be true for $n=2$ ($m=1$ and $m+1=2$). By the same logic, it is true for 2, 3, 4, etc., and therefore it is true for all $n\in\mathbb{N}$.
Here, the base case is clearly true: the left hand side is 1 and the right hand side is $\frac{2+6+5-1}{12}=1$. For the inductive step, you know that $\sum^{m+1}_{k=1} k^5=\sum^{m}_{k=1} k^5+(m+1)^5$. So you need to show that $[2(m+1)^6+6(m+1)^5+5(m+1)^4-(m+1)^2]/12=(2m^6+6m^5+5m^4-m^2)/12+(m+1)^5$. Unfortunately, this requires a lot of messy algebra.
(b): To integrate a function $f$, you need to add up all the $f(x_k)(\Delta x)_k$'s. (For each sample rectangle $k$, $f(x_k)$ is the value of $f$ somewhere in the rectangle, and $(\Delta x)_k$ is the width of the rectangle). In this case, $f(x)=x^5$ and $x_k=kb/n$, so $f(x_k)=(b/n)^5 * k^5$. You are cutting up the interval $[0,b]$ into $n$ rectangles, so the width of each one is $b/n$. The Riemann sum is $\sum^n_{k=1} f(x_k)(\Delta x)_k=\sum^n_{k=1} (b/n)^5 * k^5 * (b/n)=(b/n)^6 \sum^n_{k=1} k^5=(\frac{b}{n})^6 \frac{2n^6+6n^5+5n^4-n^2}{12}=b^6 \frac{2n^6+6n^5+5n^4-n^2}{12n^6}$. The integral is the limit of the Riemann sum as the number of sample rectangles ($n$) goes to infinity. The $n^5,n^4,n^2$ terms become negligible compared to the $n^6$ term as $n$ becomes very large, so the final answer is $\frac{1}{6}b^6$.