One can easily verify that $\|A\|=\sup\{|\langle Ax,y\rangle| : x,y \in \mathcal H,\
\|x\|=\|y\|=1\}$ in a complex Hilbert space $\mathcal H$.
Notice that $$\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle = 2\langle Ax,y\rangle+2\langle Ay,x\rangle.$$ Since $A$ is self adjoint, $\langle Ay,x\rangle=\langle y,Ax\rangle=\overline{\langle Ax,y\rangle}.$ So $$\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle=4\newcommand{\re}{\operatorname{Re}}\re\langle Ax,y\rangle.$$
Let $P:=\sup_{||x||\leq 1}|\langle x,Ax\rangle|.$ Then
$$\begin{align*}
|4\re\langle Ax,y\rangle|
&=|\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle |\\
&\leq P\|x+y\|^2+P\|x-y\|^2\\
&=2P(\|x\|^2+\|y\|^2).
\end{align*}$$
So, whenever $\|x\|=\|y\|=1$, we have $$|\re\langle Ax,y\rangle|\leq P\tag{$\color{red}{1}$}\label1.$$
Suppose $\langle Ax,y\rangle=re^{i\theta}$ with $\|x\|=\|y\|=1$. I will construct an element $z$ with $\|z\|=1$ such that $|\langle Ax,y\rangle|=|\re\langle Az,y\rangle|$. Then we can apply $(\ref 1)$ to $|\re\langle Az,y\rangle|$ and we will get $|\langle Ax,y\rangle|\leq P $.
Consider $z=e^{-i\theta} x$. Then $\|z\|=1$. Also, note that $$\langle Az,y\rangle=e^{-i\theta}\langle Ax,y\rangle=r=\re\langle Az,y\rangle,$$ and $|\langle Ax,y\rangle|=r$. So $$|\langle Ax,y\rangle|=r=|\re\langle Az,y\rangle|\leq P.$$ Hence $\|A\|\leq P$.
I learnt this proof from Functional Analysis by B.V. Limaye.
Let $H$ be the underlying Hilbert space, $I$ the identity operator on $H$, $V=\operatorname{Range}(T-I)$ and $\overline{V}$ the closure of $V$. It is a standard result of operators on Hilbert space that $V^\perp = \operatorname{Ker}((T-I)^*) = \operatorname{Ker}(T^*-I)$. Since $T^*$ is assumed to have no positive eigenvalues, $V^\perp = \operatorname{Ker}(T^*-I)=0$. Hence $V$ is dense in $H$.
For any $u\in\mathcal{D}(T)$,
$$\|(T-I)u\|^2 = \langle Tu-u,Tu-u\rangle = \|Tu\|^2+\|u\|^2 - \langle Tu,u\rangle - \langle u,Tu\rangle = \|Tu\|^2+\|u\|^2 - 2\langle u,Tu\rangle$$
where $\langle Tu,u\rangle = \langle u,Tu\rangle$ by the symmetry of $T$. Since $\langle u,Tu\rangle \le 0$ by assumption, we conclude
$$\|(T-I)u\| \ge \|u\|$$
It follows that not only is $(T-I)$ one-to-one, but it is bounded below. So considering the inverse $(T-I)^{-1}$ defined on $V$, it is a bounded linear operator. If $z\in\overline{V}$ there exists a sequence $z_n\in V$ such that $z_n\rightarrow z$. Then $z_n$ is Cauchy. Letting $y_n = (T-I)^{-1}(z_n)$, we have that $y_n$ is Cauchy because $(T-I)^{-1}$ is bounded. Let $y$ be the limit of $y_n$. then $y_n\rightarrow y$ and $z_n=(T-I)y_n \rightarrow z$. Since $T$ is closed, so is $(T-I)$. Therefore $z\in V$ and $z=(T-I)y$. Hence $\overline{V} = V$. We already showed that $V$ is dense, so $V=H$.
Now, $T$ is symmetric, so $T^*$ is an extension of $T$, so $(T^*-I) = (T-I)^*$ is an extension of $(T-I)$. But $V=\operatorname{Range}(T-I)$ is already all $H$, so either $(T^*-I)$ isn't one-to-one, or $(T^*-I) = (T-I)$. We know however from assumption that $\operatorname{Ker}(T^*-I) = 0$. So $(T^*-I)$ is one-to-one, and it follows that $(T^*-I) = (T-I)$. From this it follows that $T^*=T$.
Best Answer
Since you requested a hint, I'll provide a modest one, then put a spoiler of the solution at the end (hover your cursor over the block).
Hint: Write $a$ in terms of the orthonormal eigenbasis guaranteed by $\phi$, then just see what pops out when you compute $\frac{\langle \phi(a),a\rangle}{\langle a,a\rangle}$.
Spoiler: