Let $B(t)$, $t \ge 0$ be the standard Brownian motion. Define Brownian bridge as $$ X(t) = B(t) - tB(1), \qquad 0 \le t \le 1. $$ We see that \begin{align*} cov(X(t),B(1)) &= cov(B(t)-tB(1),B(1))\\&= cov(B(t),B(1)) - tcov(B(1),B(1)) = t - t = 0. \end{align*} Hence, since $X(t)$ is a Gaussian process, it is independent of $B(1)$.
Now my question:
Let $0 < t_1 < t_2 < \dots < t_n < 1$. Is the vector $$ (X(t_1),X(t_2),\dots,X(t_n)) $$ independent of $B(1)$?
Yes. Since the variable are jointly normal the fact that $cov(X(t_i),B(1))=0$ for each $i$ implies that $(X(t_1),X(t_2),...,X(t_n))$ is independent of $B(1)$.