Well, for an arbitrary inner product on $X:=H^*$ it is not going to work, since then $X^*$ need not be isomorphic to $H$.
On the other hand, the Riesz representation gives a linear isomorphism $H\to H^*$, and if the inner product is defined via this isomorphism, i.e. if
$$(\langle x,-\rangle,\langle y,-\rangle)_{H^*}=\langle x,y\rangle_{H} $$
for all $x,y\in H, \ f\in H^*$, then your claim is valid:
Let $a,b\in H^*$ then $a=\langle x,-\rangle$ and $b=\langle y,-\rangle$ for some $y\in H$ by Riesz representation, and $a(y)=\langle x,y\rangle$ so we have
$$(a,b)_{H^*} = (a,\langle y,-\rangle)=a(y)=\langle a,y\rangle_{H^*,H}\ .$$
Just a hint:
$\phi(p +aq) = p(1) + aq(1) = \phi(p) + a\phi(q)$ therefore $\phi$ is linear.
$||\phi|| = \sup_{p : \int_0^1p^2 = 1}\phi(p) = \sup_{p : \int_0^1p^2 = 1}p(1)$.
$p = ax^2 + bx +c$, $\int_0^1p^2 = 1 \Longleftrightarrow \frac{a^2}{5} + \frac{ab}{2} + \frac{2ac+b^2}{3} + bc + c^2 = 1$.
Now you have to maximize $p(1) = a +b + c$ bearing in mind the previous condition.
Also: $p = ax^2 + bx +c$, $q = dx^2 + ex +f$.
$\int_0^1pq = a(\frac{d}{5} + \frac{e}{4} + \frac{f}{3}) + b(\frac{d}{4} + \frac{e}{3} + \frac{f}{2}) + c(\frac{d}{3} + \frac{e}{2} + f)$.
You are looking for $q$ such that the previous integral equals $p(1) = a + b +c$, so you look for $d,e,f$ such that $\frac{d}{5} + \frac{e}{4} + \frac{f}{3} = 1$, $\frac{d}{4} + \frac{e}{3} + \frac{f}{2} = 1$ and $\frac{d}{3} + \frac{e}{2} + f = 1$.
Best Answer
The map $f$ is clearly injective, and an isometry. Suppose it is surjective : then $H$ is isometric to its dual space. For any normed vector space $X$ and Banach space $Y$, the space of continuous linear maps $$\mathrm{L}(X,Y)$$ equipped with the usual norm is automatically a Banach space, in particular for any normed vector space $X$, its topological dual $X'$ is a Banach space. Thus, $H$ is isometric to a Banach space, complete and so $H$ is a Hilbert space.