= Solution
First the <eigenvalues> of <Hecke operators> are real. Here is the needed <Petersson inner product> argument: unfold the finite coset sum defining $T_n$ into its correspondence domain, then change variables by each matrix. The determinant-normalized slash action preserves $y^k\,dx\,dy/y^2$ in the paired integrand, and transfers a representative to its inverse. Multiplying that inverse by the positive scalar $n$ gives its adjugate without changing the slash action. Adjugation reverses the integer determinant-$n$ correspondence, so the resulting sum is again $T_n$. Consequently $\langle T_nf,g\rangle=\langle f,T_ng\rangle$. Taking $g=f$ proves reality of $\lambda(n)$, since $\langle f,f\rangle>0$.
Fix one prime $p$ and set $x=\lambda(p)/p^{(k-1)/2}$ and $b_e=\lambda(p^e)/p^{e(k-1)/2}$. The <Hecke multiplication relations> proved in (a) give $b_0=1$, $b_1=x$ and $b_{e+2}=xb_{e+1}-b_e$. The distinct-root assumption excludes $x=\pm2$.
If $|x|<2$, write $x=2\cos\theta$, $0<\theta<\pi$. Solving the <linear recurrence> gives $b_e=\sin((e+1)\theta)/\sin\theta$. If $\theta/(2\pi)$ is rational, infinitely many $e$ satisfy $(e+1)\theta\equiv\theta\pmod{2\pi}$, hence $b_e=1$. If it is irrational, density of the corresponding <irrational rotation> gives infinitely many $e$ in the open arc where $\sin((e+1)\theta)>\sin\theta/2$.
If $|x|>2$, let $\beta,\beta^{-1}$ be the real roots of $X^2-xX+1$, choosing $|\beta|>1$. Then $b_e=(\beta^{e+1}-\beta^{-e-1})/(\beta-\beta^{-1})$. Its even-index terms are positive and tend to infinity, whether $\beta$ is positive or negative. Thus the <prime-power coefficients of a level-one eigenform are often large and positive>, and in either case
$$
\boxed{\lambda(p^e)\ge\tfrac12p^{e(k-1)/2}\text{ for infinitely many }e.}
$$
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