= Solution
{c}
We use the equilibrium <generalized Langevin equation> and its <generalized fluctuation-dissipation relation>, with the factor $5$ in the noise <autocorrelation> fixing the equilibrium <velocity> <variance>. Let $g$ be the <velocity> relaxation function and let a tilde denote a one-sided <Laplace transform>:
$$
g'(t)=-\int_0^t\kappa(u)g(t-u)\,du,
\qquad g(0)=1,
\qquad
\widetilde g(s)=\frac1{s+\widetilde\kappa(s)}.
$$
The equilibrium <velocity> <autocorrelation function> is $\chi(t)=5g(t)$, so
$$
\boxed{\widetilde\chi(s)=\frac5{s+\widetilde\kappa(s)},\qquad \chi(0)=5.}
$$
For completeness, this relation follows without assuming independence of the noise and <velocity> in a stationary infinite history. The two-sided noise <power spectral density> is $S_R(\omega)=10\operatorname{Re}\widetilde\kappa(i\omega)$. The stable linear equation has <velocity> transfer function $H(i\omega)=[i\omega+\widetilde\kappa(i\omega)]^{-1}$, so
$$
S_V(\omega)=|H(i\omega)|^2S_R(\omega)
=10\operatorname{Re}H(i\omega).
$$
This is precisely the <power spectral density> of the even extension of $5g(t)$. Stable relaxation removes initial transients with a finite second <moment>, giving the long-time <autocorrelation function> defined in the question. All the parameter choices used below have stable relaxation.
Integrating the <velocity autocorrelation function> twice gives the <mean squared displacement> from a fixed initial <position>:
$$
\mathbb E\bigl[(X(t)-X(0))^2\bigr]
=2\int_0^t(t-u)\chi(u)\,du.
$$
A decaying initial transient does not change its asymptotic slope. Therefore the <Green-Kubo formula for diffusion> gives
$$
D=\int_0^\infty\chi(u)\,du
=\widetilde\chi(0)
=\frac5{\widetilde\kappa(0)}.
$$
For the given friction <memory kernel>, direct <integration> yields
$$
\widetilde\kappa(s)=\frac{c_1}{s+c_3}+\frac{c_1c_2}{(s+c_3)^2},
\qquad
\widetilde\kappa(0)=\frac{c_1}{c_3}+\frac{c_1c_2}{c_3^2}.
$$
At $c_1=c_2=c_3=1$, the integrated friction is $2$, hence
$$
\boxed{D=\frac52.}
$$
The relaxation denominator here is $s^3+2s^2+2s+2$. Its roots have negative real parts by the cubic <Routh-Hurwitz criterion>, since all coefficients are positive and $2\cdot2>2$. This justifies the decay and <integrability> used above. The <diffusion coefficient> depends on the area under the friction <memory kernel>, not just its value at zero.
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