= Solution
By part (a), translated by $a$, starting with $B_0\sim\nu_{a,R}$ gives
$$
B_{\tau_{a,r_1}}\sim\nu_{a,r_1}.
$$
Part (c) says that the same hitting law from $\nu_{0,R}$ tends in <total variation distance> to $\nu_{a,r_1}$. On the other hand, every path from the circle of radius $R>r_2$ to the target disc must first hit the circle of radius $r_2$. Part (a) and the <Strong Markov property> show that its position there has law $\nu_{0,r_2}$ and that
$$
\mathbb P_{\nu_{0,R}}(B_{\tau_{a,r_1}}\in A)
=\mathbb P_{\nu_{0,r_2}}(B_{\tau_{a,r_1}}\in A),
$$
independently of $R$. Letting $R\to\infty$ in part (c) proves
$$
B_{\tau_{a,r_1}}\sim\nu_{a,r_1}
$$
when $B_0\sim\nu_{0,r_2}$.
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