Mean and covariance of Wiener process Let $x(t), x(0)=0$ be a Wiener process with the parameters $a$ and $\sigma.$ Prove  that  its mean equals $a \cdot t$ and its covariance $R(t,s)$ is equal $R(t,s)=\sigma \min(t,s)$

When you say 

Let $x(t), x(0)=0$ be a Wiener process with the parameters $a$ and $\sigma$

you presumably mean that $x(t)=at+\sigma W(t)$ where $W(t)$ is a standard Wiener process, i.e. $W(0)=0$, $E[W(t)] =0$ and $W(t)-W(s) \sim N(0, t-s)$ when $t \gt s$. 
Then $E[x(t)]= aE[t] + \sigma E[W(t)] = at+0=at $.
Assume  $s$ is the minimum of the two, so $0 \lt s \lt t$.  Then $W(t)-W(s)$ and $W(s)-W(0)$ are independent with zero means so $E[(W(t)-W(s))(W(s)-W(0))]=0$ so $E[W(t)W(s)]=E[W(s)^2]=Var(W(s))=s$.  
So $Cov(x(t), x(s)) =E[(x(t)-E[x(t)])(x(s)-E[x(s)])] = E[(\sigma W(t))(\sigma W(s))] = \sigma^2 s$.  This squaring of $\sigma$ is not quite what your question asked to prove, so I suspect there might be an error in the question. 
