(Proof Updated, Verification in Need) Show that $(s,t)\mapsto t\wedge s$ for $s,t\geq 0$ and $(s,t)\mapsto e^{-|t-s|}$ are positive semi-definite. We say a positive symmetric $n\times n$ matrix $M$ over $\mathbb{R}^n$ is semi-definite if $v^{\intercal}Mv\geq 0$ for all nonzero $v\in\mathbb{R}^n$.
We say a function $f:\mathbb{T}^2\longrightarrow\mathbb{R}$ to be positive semi-definite if $\Big(f(t_k, t_j)\Big)_{k,j=1}^n$ is a positive semi-definite matrix for all $(t_k)_{k=1}^n\in\mathbb{T}^n$
With this definition, I am working on an exercise asking me to show

$(1)$ the function $f:(s,t)\mapsto t\wedge s$ defined for $s,t\geq 0$ is positive semi-definite;
$(2)$ the function $c:(s,t)\mapsto e^{-|t-s|}$ is positive semi-definite.

For the first one, I tried to use the fact that $$\int_{\mathbb{R}^{+}}\mathbb{1}_{[0,t]}\mathbb{1}_{[0,s]} \,d\mu = t\wedge s,$$ so that each term in the matrix is of the form $b_{i,j}:=f(t_i,t_j)=\int_{\mathbb{R}^{+}}\mathbb{1}_{[0,t_{j}]} \mathbb{1}_{[0,t_{}} \, d\mu$, for $t,j=1,\cdots,n$.
Let $v=(v_1,\ldots, v_n)\in\mathbb{R}^n$, then $$v^{\intercal} Mv=a_1 \sum_{i=1}^n a_i b_{i,1}+a_2\sum_{i=1}^n a_i b_{i,2} + a_3 \sum_{i=1}^n a_i b_{i,3}+\cdots+a_n \sum_{i=1}^n a_i b_{i,n}.$$
But then I don't know what to do next..
For the second one, the exercise gives an hint: using an auxiliary Hilbert space $H$ and a $h_t\in H$ such that $\langle h_t, h_s\rangle=c(s,t)\ldots$ I don't really know how to use this hint...
I really need an answer with some details, since this is an exercise in Stochastic Process, instead of functional analysis and so forth, so I don't have enough background of this...
Thank you so much!

Edit 1: (Proof of the first one)
Following MaoWao's suggestion, I think I proved the first one.
Firstly let me claim that if $H$ is a Hilbert space, then its corresponding inner product $\langle\cdot, \cdot\rangle_H:H\times H\longrightarrow\mathbb{R}$ is positive semi-definite. 
Indeed, we have for any $n\in\mathbb{N}$, $x_1,\ldots, x_n \in H$ and $c_1,\ldots, c_n \in\mathbb{R}$ that $$\sum_{i,j=1}^n c_i c_j \langle x_i,x_j\rangle_H = \left<\sum_{i=1}^n c_i x_i,\sum_{j=1}^n c_j x_j \right>_H =\Big\|\sum_{i=1}^n c_i x_i\Big\|_H^2\geq 0.$$
In fact, the above result also holds for pre-Hilbert space, since the notion of completeness was not involved in the above argument. 
Thus, we only need to find a specific (pre-)Hilbert Space $H$ and a $h_{t}\in H$ such that $\langle h_t, h_s\rangle_H = t\wedge s$.
But this is easy, let's consider $H:=L^2(\mathbb{R}_{+})$, and $h_t:=\mathbb{1}_{[0,t]}$. It is clear that $h_t\in H$, and for any $t,s\geq 0$, we have $$\langle h_t, h_s\rangle_{L^2(\mathbb{R}_{+})} = \int_{\mathbb{R}_{+}}\mathbb{1}_{[0,t]} \mathbb{1}_{[0,s]} \, d\mu=t\wedge s,$$ and thus we are done. 

Edit 2: (Proof of the second one)
I've searched all over the places. The function in $(2)$ is Abel kernel, but it is rarely discussed since Abel kernel is closely related to Poisson kernel, and most of the discussions are on the latter. 
Later, I found a really close one: the Gaussian kernel and here is a link about the proof of Gaussian kernel is really a kernel. That is, it is positive semidefinite. 
https://stats.stackexchange.com/questions/35634/how-to-prove-that-the-radial-basis-function-is-a-kernel
In this link, one answer used the characteristic function. This greatly inspired me. I also found a characteristic function, which is the one for Cauchy distribution. 
Below is the proof:
Recall the Cauchy Distribution $(x_{0},\gamma)$ with $\gamma>0$ has the characteristic function $$\varphi(t)=e^{ix_{0}t-\gamma|t|}.$$
Using this, we can write $c(s,t)=h(s-t)$ where $h(t):=e^{-|t|}=\mathbb{E}e^{itZ}$ is the characteristic function of a random variable $Z$ with Cauchy $(0,1)$ distribution.
Then for real numbers $x_{1},\cdots, x_{n}$ and $a_{1},\cdots, a_{n}$, we have 
\begin{align*}
\sum_{j,k=1}^{n}a_{j}a_{k}h(x_{j}-x_{k})&=\sum_{j,k=1}^{n}a_{j}a_{k}\mathbb{E}e^{i(x_{j}-x_{k})Z}\\
&=\mathbb{E}\Big(\sum_{j,k=1}^{n}a_{j}e^{ix_{j}Z}a_{k}e^{-ix_{k}Z}\Big)\\
&=\mathbb{E}\Big(\Big|\sum_{j=1}^{n}a_{j}e^{ix_{j}Z}\Big|^{2}\Big)\geq 0.
\end{align*}
Thus, $c$ is positive semi-definite.
It seems that I did not use the hint at all for part $(2)$, so I believe there must be another way.
I do need someone to check if my proof in the edit 1 and edit 2 is correct. I am gonna open a bounty in 19 hours later, for proof checking and possible new proof. Thank you!
 A: As MaoWao suggested, I am gonna post my own proof here.

Proof of (1):
Firstly let me claim that if $H$ is a Hilbert space, then its corresponding inner product $\langle \cdot, \cdot\rangle_{H}:H\times H\longrightarrow\mathbb{R}$ is positive semi-definite. 
Indeed, we have for any $n\in\mathbb{N}$, $x_{1},\cdots, x_{n}\in H$ and $c_{1},\cdots, c_{n}\in\mathbb{R}$ that $$\sum_{i,j=1}^{n}c_{i}c_{j}\langle x_{i}, x_{j}\rangle_{H}=\Big<\sum_{i=1}^{n}c_{i}x_{i},\sum_{j=1}^{n}c_{j}x_{j}\Big>_{H}=\Big\|\sum_{i=1}^{n}c_{i}x_{i}\Big\|_{H}^{2}\geq 0.$$
In fact, the above result also holds for pre-Hilbert space, since the notion of completeness was not involved in the above argument. 
Thus, we only need to find a specific (pre-)Hilbert Space $H$ and a $h_{t}\in H$ such that $\langle h_{t}, h_{s}\rangle_{H}=t\wedge s$.
But this is immediate. Let's consider $H:=L^{2}(\mathbb{R}_{+})$, and $h_{t}:=\mathbb{1}_{[0,t]}$. It is clear that $h_{t}\in H$, and for any $t,s\geq 0$, we have $$\langle h_{t}, h_{s}\rangle_{L^{2}(\mathbb{R}_{+})}=\int_{\mathbb{R}_{+}}\mathbb{1}_{[0,t]}\mathbb{1}_{[0,s]}d\mu=t\wedge s,$$ and thus we are done. 

Proof of $(2)$:
Recall the Cauchy Distribution $(x_{0},\gamma)$ with $\gamma>0$ has the characteristic function $$\varphi(t)=e^{ix_{0}t-\gamma|t|}.$$
Using this, we can write $c(s,t)=h(s-t)$ where $h(t):=e^{-|t|}=\mathbb{E}e^{itZ}$ is the characteristic function of a random variable $Z$ with Cauchy $(0,1)$ distribution.
Then for real numbers $x_{1},\cdots, x_{n}$ and $a_{1},\cdots, a_{n}$, we have 
\begin{align*}
\sum_{j,k=1}^{n}a_{j}a_{k}h(x_{j}-x_{k})&=\sum_{j,k=1}^{n}a_{j}a_{k}\mathbb{E}e^{i(x_{j}-x_{k})Z}\\
&=\mathbb{E}\Big(\sum_{j,k=1}^{n}a_{j}e^{ix_{j}Z}a_{k}e^{-ix_{k}Z}\Big)\\
&=\mathbb{E}\Big(\Big|\sum_{j=1}^{n}a_{j}e^{ix_{j}Z}\Big|^{2}\Big)\geq 0.
\end{align*}
Thus, $c$ is positive semi-definite.

Also, as MaoWao pointed out, the proof of $(2)$ can be equivalently written into something similar to what we did in the $(1)$, by taking the Hilbert space $L^{2}(\mathbb{P})$ and $h_{t}=\exp(it\cdot)$
I am really grateful to the help, suggestion and proof verification from MaoWao. I'd like also to express my appreciation to Michael Hardy, for the discussion and help. 
Thank you guys :)
A: This will be a partial answer.
The standard Wiener process $\{B(t)\}_{t\,\ge\,0},$ also called the standard Brownian motion, assigns to each $t\ge0$ a random variable $B(t)$ in such a way that every increment $B(t)-B(s)$ for $0\le s\le t$ is distributed as $\operatorname N(0,t-s)$ (the standard normal distribution with expected value $0$ and variance $t-s,$ so standard deviation $\sqrt{t-s},$ and for pairwise disjoint intervals $(s_i,t_i),$ $i=1,\ldots,n,$ with $0\le s_i\le t_i,$ the increments $B(t_i) - B(s_i)$ are independent random variables.
Then for $0\le s\le t$ one can show that
$$
\operatorname{cov}(B(s), B(t)) = s = s\wedge t.
$$
That is done by writing
$$
\operatorname{cov}\Big(B(s), \Big(B(t) - B(s)\Big) + B(s)\Big)
$$
and then using the independence of $B(t) - B(s)$ and $B(s).$
Since all the entries in the matrix of covariances of $B(t_1),\ldots, B(t_n)$ are of the form $t_i\wedge t_j,$ and since every matrix of covariances is positive-semi-definite, the function $(s,t)\mapsto s\wedge t$ must be positive-semi-definite.
