= Solution
Use the product printed in the original PDF, with factors $(1-q^n)^2$ and $(1-q^{11n})^2$. The <Fricke involution> normalizes $\Gamma_0(11)$ and preserves its <modular cusp> space. Therefore $g=-11^{-1}\tau^{-2}f(-1/(11\tau))$ lies in the given one-dimensional space, so $g=cf$ for a constant $c$.
At the Fricke fixed point $\tau_*=i/\sqrt{11}$, the prefactor $-11^{-1}\tau_*^{-2}$ is one. Thus $g(\tau_*)=f(\tau_*)$. The product has $0<q_*<1$, every factor is positive, and its limit is nonzero since $\sum_nq_*^n+\sum_nq_*^{11n}<\infty$. Hence $f(\tau_*)>0$, forcing $c=1$. This <Fricke sign from a nonvanishing fixed-point value> proves
$$
\boxed{-\frac1{11}\tau^{-2}f\left(-\frac1{11\tau}\right)=f(\tau).}
$$
Part (b) now gives $\Lambda(f,s)=\Lambda(f,2-s)$. Its Taylor series at one contains only even powers, so its order of vanishing is even. The function is not identically zero, since its first <Fourier coefficient> is one. In this example the product is positive on the entire positive imaginary axis, and its Mellin integral at $s=1$ is positive. Thus the stronger conclusion is
$$
\boxed{\operatorname{ord}_{s=1}\Lambda(f,s)=0,\quad\text{in particular it is even}.}
$$
The TeX aid duplicates and corrupts the product in this part; neither corrupted expression is used.
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