An inner product, such as you have here, if a way to measure the angle between two vectors. In this case a "vector" is a function on the interval $[-\pi,\pi]$. You put two of these into the operation, and get a number out. If that number is zero, your vectors are called orthogonal or perpendicular. If not, it measures how close to orthogonal they are, based on the formula $$\cos \theta=\frac{\langle u,v\rangle}{\sqrt{\langle u,u\rangle}\sqrt{\langle v,v\rangle}}$$
For a concrete example, let $u=x, v=x^2$. Now $$\langle u,v\rangle=\frac{1}{\pi}\int_{-\pi}^\pi uv\;dx=\frac{1}{\pi}\int_{-\pi}^\pi x^3\;dx=\left.\frac{1}{4\pi}x^4\right|_{-\pi}^{\pi}=0$$
Hence $u,v$ are orthogonal under this inner product.
Okay, now to the questions at hand. You are tasked with proving that (I) $V$ is indeed a vector space, and that (II) $\langle \cdot,\cdot\rangle$ as defined is indeed an inner product. Each of these requires verifying a long list of little properties, many of which seem obvious; however they only appear so because the examples here are familiar.
Part (I): Here are the properties you need to check. The most basic are closure: namely that the sum of two continuous functions is continuous (vector addition), and that the product of a continuous function by a real number is continuous (scalar multiplication). But, there are also ones like the sum of two continuous functions is the same regardless of the order in which you add. This is hard to check unless you're given a precise definition of $\S^1$ (as you call it).
Part (II): Here are the properties you need to check. Since you're dealing with real (not complex) functions, this simplifies slightly to five properties:
For all continuous functions $u,v$, $\langle u,v\rangle=\langle v,u\rangle$.
For all continuous functions $u,v$ and all real numbers $a$, $\langle au,v\rangle=a\langle u,v\rangle$.
For all continuous functions $u,v,w$, $\langle u+v,w\rangle=\langle u,w\rangle+\langle v,w\rangle$
For all continuous functions $u\neq 0$, $\langle u,u\rangle > 0$
For the continous function $u=0$, $\langle u,u\rangle=0$.
For each of these, you need to use the definition of $\langle \cdot,\cdot\rangle$ that you were given; the properties you need (except for #4) sort of fall out from the familiar properties of integrals you already know.
$\langle u,v\rangle=\frac{1}{\pi}\int_{-\pi}^\pi u(x)v(x)\,dx=\frac{1}{\pi}\int_{-\pi}^\pi v(x)u(x)\,dx=\langle v,u\rangle$.
$\langle au,v\rangle=\frac{1}{\pi}\int_{-\pi}^\pi au(x)v(x)\,dx=\frac{a}{\pi}\int_{-\pi}^\pi u(x)v(x)\,dx=a\langle u,v\rangle$.
$\langle u+v,w\rangle=\frac{1}{\pi}\int_{-\pi}^\pi (u(x)+v(x))w(x)\,dx=\frac{1}{\pi}\int_{-\pi}^\pi u(x)w(x)\,dx+\frac{1}{\pi}\int_{-\pi}^\pi v(x)w(x)\,dx=\langle u,w\rangle+\langle v,w\rangle$
$\langle u,u\rangle=\frac{1}{\pi}\int_{-\pi}^\pi u(x)^2\,dx$ (need to prove this is $>0$)
$\langle u,u\rangle=\frac{1}{\pi}\int_{-\pi}^\pi 0^2\,dx=0$
Now, for #4, there is something to prove; namely that every continuous function except 0 has a positive integral. This is called the mean value theorem for integrals of continuous functions; you need to cite it or prove it.
Best Answer
$$\langle p, q \rangle = \langle x^2-x + 1/6, x^2-x + 1/6 \rangle$$
$$ = \int_{0}^{1}(x^2-x + 1/6)^2dx$$ $$ = \int_{0}^{1}\left(x^4-2x^3 + \frac43x^2 - \frac13x+ \frac{1}{36}\right)dx$$