= Solution
Write $u(t)=e^t$ and $v(t)=e^{-t}$. Both lie in the <Hilbert space> $L^2(-\pi,\pi)$, and the <integral operator> factors as the <rank-one operator>
$$
Tf=\langle f,v\rangle u.
$$
The <Cauchy-Schwarz inequality> gives $\|Tf\|\leq\|u\|\|v\|\|f\|$, and equality is attained at $f=v/\|v\|$. Hence the <operator norm> is $\|T\|=\|u\|\|v\|$. Direct integration gives
$$
\|u\|^2=\int_{-\pi}^{\pi}e^{2t}\,dt
=\frac{e^{2\pi}-e^{-2\pi}}2=\sinh(2\pi),\qquad
\|v\|^2=\sinh(2\pi).
$$
For any <Hilbertian basis> $(h_k)$, the <Parseval identity for a Hilbertian basis> gives
$$
\|T\|_2^2=\sum_k\|\langle h_k,v\rangle u\|^2
=\|u\|^2\sum_k|\langle h_k,v\rangle|^2
=\|u\|^2\|v\|^2.
$$
This is the general formula for the <norms of a rank-one operator>. In particular $T$ is a <Hilbert-Schmidt operator>, and
$$
\boxed{\|T\|=\|T\|_2=\sinh(2\pi)=\frac{e^{2\pi}-e^{-2\pi}}2.}
$$
This computation needs no explicit <Fourier basis>. The printed optional basis formula omits the variable $t$: the intended <Hilbertian basis> is $e_n(t)=(2\pi)^{-1/2}e^{int}$, $n\in\mathbb Z$. Taken literally, the printed expressions are constant <functions> and cannot form an <orthonormal basis> of $L^2(-\pi,\pi)$.
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