We define a sequence of (discrete) stopping times
$$\tau_j := \frac{\lfloor 2^j \tau \rfloor+1}{2^j}, \qquad j \in \mathbb{N}.$$
It is not difficult to see that $\tau_j$ is indeed a stopping time and $\tau_j \downarrow \tau$ as $j \to \infty$. Since the Brownian motion has continuous paths, this implies $B(\tau) = \lim_{j \to \infty} B(\tau_j)$.
Let $\xi,\eta \in \mathbb{R}$. Then, by the dominated convergence theorem,
$$\begin{align*} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg) &= \lim_{j \to \infty} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau_j+t)-B(\tau_j))} \cdot e^{\imath \, \eta B(\tau_j)} \bigg) \\ &= \lim_{j \to \infty} \sum_{k=1}^{\infty} \mathbb{E} \bigg( e^{\imath \, \xi (B(k \cdot 2^{-j} +t)-B(k \cdot 2^{-j}))} \cdot e^{\imath \, \eta B(k \cdot 2^{-j})} \cdot 1_{\{\tau_j = k \cdot 2^{-j}\}} \bigg) \end{align*}$$
where we used in the last step that $\tau_j$ is a discrete stopping time. By assumption, $B(k \cdot 2^{-j}+t)-B(k \cdot 2^{-j})$ and $B(k \cdot 2^{-j}) \cdot 1_{\{\tau_j=k2^{-j}\}}$ are independent. Therefore, we obtain
$$\begin{align*} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg) &= \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg) \lim_{j \to \infty} \sum_{k=1}^{\infty} \mathbb{E} \bigg( e^{\imath \, \eta B(k 2^{-j})} \cdot 1_{\{\tau_j=k \cdot 2^{-j}\}} \bigg) \\ &= \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg) \cdot \mathbb{E}\bigg(e^{\imath \, \eta B(\tau)} \bigg). \end{align*}$$
(In the second step we used again dominated convergence, similar to the above calculation.) If we choose $\eta = 0$, then we get
$$ \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \bigg) = \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg);$$
hence,
$$ \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg)= \mathbb{E}\bigg(e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) \cdot \mathbb{E}\bigg(e^{\imath \, \eta B(\tau)} \bigg)$$
i.e. $B(\tau+t)-B(\tau)$ and $B(\tau)$ are independent. Therefore, the strong Markov property gives
$$\begin{align*} \mathbb{E}\bigg(1_F e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) &= \mathbb{E}\bigg(1_F \mathbb{E} \bigg[ e^{\imath \, \xi (B(\tau+t)-B(\tau))} \mid \mathcal{F}_{\tau} \bigg] \bigg) \\ &= \mathbb{E}\bigg(1_F \mathbb{E} \bigg[ e^{\imath \, \xi (B(\tau+t)-B(\tau))} \mid B_{\tau} \bigg] \bigg)\\ &= \mathbb{P}(F) \cdot \mathbb{E}\bigg(e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) \end{align*}$$
for any $F \in \mathcal{F}_{\tau}$. Consequently, $B(\tau+t)-B(\tau)$ is independent of $\mathcal{F}_{\tau}$.
A very similar calculation shows that
$$\mathbb{E} \left( \exp \left( \imath \sum_{j=1}^n \xi_j \cdot (B(\tau+t_j)-B(\tau+t_{j-1})) \right) \right) = \prod_{j=1}^n \mathbb{E}e^{\imath \, \xi_j B(t_j-t_{j-1})}$$
for any $\xi_j \in \mathbb{R}$, $0 \leq t_0 < \ldots \leq t_n$. This means that $(B(\tau+t_j)-B(\tau+t_{j-1}))_{j=1,\ldots,n}$ are independent normal distributed random variables.
It is true that the second property can be deduced from the first one. Indeed, the first property implies that $(B_t)$ has stationary, independent increments. Hence, the following statement would give you the implication in the forward direction:
Proposition: Any real-valued stochastic process with stationary, independent increments has the elementary Markov Property.
Proof: We just need to show that $E[f(X_t)|\mathcal F_s] = E[f(X_t)|X_s]$ for any $s<t$ and any bounded, measurable function $f: \Bbb R \to \Bbb R$. It's straightforward to show that this is equivalent to the above definition.
Notice that if $t>s$, then for any (bounded-measurable) function $f: E \to \mathbb{R}$, we can write $$E\big[f(X_t)\big|\mathcal{F}_s\big] = E\big[f(X_t-X_s+X_s)\big|\mathcal{F}_s\big] = E\big[g(X_t-X_s,X_s)\big|\mathcal{F}_s\big]$$ where $g(x,y) := f(x+y)$. We know that $X_s$ is $\mathcal{F}_s$-measurable, and that $X_t-X_s$ is independent of $\mathcal{F}_s$ (this is just independence of increments). Therefore, using this question (Conditional Expectation of Functions of Random Variables satisfying certain Properties), we know that $$E\big[g(X_t-X_s,X_s)\big|\mathcal{F}_s\big] = E\big[g(X_t-X_s,X_s)\big|X_s\big] = E\big[f(X_t)\big|X_s\big]$$It follows that $(X_t)_t$ is a Markov Process. $\Box$
Best Answer
From $2$ to $1$:
Let $V=(V_t) = (B_{T+t} - B_t)$ be that process, let $A \in \mathcal B(\mathbb R^{[0,\infty)})$ (in cylinder $\sigma-$field). Let $A_0 = \{ x \in \mathbb R^{[0,\infty)} : x-x(0) \in A \}$ ( we translate every function in $A$ by minus it's value at $0$). Note that:
$$ \mathbb P_x( V \in A | \mathcal F(T)) = \mathbb P_x( (B_{T+t} - B_T) \in A | \mathcal F(T)) = \mathbb P_x ( (B_{T+t}) \in A_0 | \mathcal F(T))$$ (since $B_T$ is value at $0$ of process $(B_{T+t})_{t \ge 0}$. Now apply $2$, getting: $$ \mathbb P_x ( (B_{T+t}) \in A_0 | \mathcal F(T)) = \mathbb P_{B_T}( (B_t) \in A_0) $$
Note that for any $y \in \mathbb R$ we have: $$ \mathbb P_y( (B_t) \in A_0 ) = \mathbb P_y( (B_t - y) \in A) = \mathbb P_0 ( (B_t) \in A) $$
Taking $y = B_T$ it finally gives us:
$$ \mathbb P_x( V \in A | \mathcal F(T)) = \mathbb P_0 ( (B_t) \in A) $$
in particular $$ \mathbb P_x (V \in A) = \mathbb E_x[\mathbb P_x(V \in A |\mathcal F(T))]= \mathbb P_0( ( B_t) \in A) $$ so we showed that $V$ has the same distribution (under $\mathbb P_x$ measure) as standard brownian motion $(B_t)$ (cause it's under $\mathbb P_0$ measure))
Now to show independence, Take any $B \in \mathcal F(T)$ we get:
$$ \mathbb P_x(B \cap \{ V \in A\}) = \mathbb E_x [ 1_B \mathbb P_0 ( (B_t) \in A)) = \mathbb P_x(B)\mathbb P_0( (B_t) \in A) = \mathbb P_x(B)\mathbb P_x( V \in A) $$
From 1 to 2:
$$\mathbb E_x [ f(B_{T+t}) | \mathcal F(T)] = \mathbb E_x [ f(B_{T+t} - B_{T} + B_{T}) | \mathcal F(T)] $$
I don't know what information you possess, but it can be shown that as adapted, right continuous process, Brownian Motion is progresivelly measurable, hence $B_T$ is $\mathcal F(T)$ measurable. By 1. we have that $B_{T+t} - B_T$ is independent of $\mathcal F(T)$ so by conditional expected value property, the last one is equal to: $$ \mathbb E_x[ f(B_{T+t} - B_T + p)] |_{p = B_T} $$ Again using $1$, we know that $B_{T+t} - B_T$ under $\mathbb P_x$ is distributed as standard (so under $\mathbb P_0$) brownian motion, so: $$ \mathbb E_x[ f(B_{T+t} - B_T + p)] |_{p = B_T} = \mathbb E_0 [ f(B_t + p)]|_{p = B_T} = \mathbb E_p[f(B_t)]|_{p =B_T} = \mathbb E_{B_T}[f(B_t)]$$
So we proved $\mathbb E_x [ f(B_{T+t}) | \mathcal F(T)] = \mathbb E_{B_T}[f(B_t)]$.