Finding the auto-correlation sequence $r_{xx}[k]$ for an AR(2) process
We have
\begin{equation}
x[n] = v[n] + 0.75x[n-1]-0.25x[n-2]
\end{equation}
You started correct, you get
\begin{equation}
 \begin{split}
  r_{xx}[k] = E\{(v[n] + 0.75x[n-1]-0.25x[n-2])(v[n-k] + 0.75x[n-k-1]-0.25x[n-k-2])\} \\
 \end{split}
\end{equation}
You get 9 terms
\begin{equation}
 \begin{split}
  r_{xx}[k] &amp;= Ev[n]v[n-k] +0.75E x[n-k-1]v[n] - 0.25E x[n-k-2]v[n]\\
   &amp;+0.75 Ex[n-1]v[n-k] + 0.75^2Ex[n-1]x[n-k-1]+0.75(0.25)Ex[n-1]x[n-k-2]\\
   &amp;-0.25Ex[n-2]v[n-k] -0.75(0.25)Ex[n-2]x[n-k-1] + 0.25^2Ex[n-2]x[n-k-2] 
 \end{split}
 \tag{1}
\end{equation}
Since the noise is white then
\begin{equation}
 Ev[n]v[n-k]  = \delta(k)\sigma_v^2 = \delta(k)
\end{equation}
Since $k\geq 0$, write down $x[n-k-1]$ and $x[n-k-2]$ do not contain lags that align with $v[n]$, so 
\begin{align}
 E x[n-k-1]v[n] &amp;= 0\\
 E x[n-k-2]v[n] &amp;= 0
\end{align}
On the other hand, $Ex[n-1]v[n-k]$ and $Ex[n-2]v[n-k]$, will agree for $k = 1$ and $k=2$ according to 
\begin{align}
x[n-1]v[n-k] &amp;= \overbrace{Ev[n-1]v[n-k]}^{\delta(k-1)} + 0.75\overbrace{ Ex[n-2]v[n-k] }^{\delta(k-2)}- 0.25\overbrace{Ex[n-3]v[n-k]}^{0} \\
x[n-2]v[n-k] &amp;= \underbrace{Ev[n-2]v[n-k]}_{\delta(k-2)} + \underbrace{E(0.75x[n-3]-0.25x[n-4])v[n-k]}_{0}
\end{align}
So
\begin{align}
 x[n-1]v[n-k] &amp;= \delta(k-1) + 0.75\delta(k-2) \\
 x[n-2]v[n-k] &amp;= \delta(k-2)
\end{align}
The other four terms follow from the AR definition, i.e.
\begin{align}
 E x[n-1]x[n-k-1] &amp;= r_{xx}[n-1-n+k+1]= r[k] \\
 E x[n-1]x[n-k-2] &amp;= r_{xx}[n-1-n+k+2]= r[k+1] \\
 E x[n-2]x[n-k-1] &amp;= r_{xx}[n-2-n+k+1]= r[k-1] \\
 E x[n-2]x[n-k-2] &amp;= r_{xx}[n-2-n+k+2]= r[k] 
\end{align}
Replacing all results in equation $(1)$, we get
\begin{equation}
\begin{split}
  r[k] &amp;= \delta(k) +0.75\delta(k-1) + 0.75^2\delta(k-1) -0.25\delta(k-2)\\
   &amp;+0.75^2 r[k] +0.75(0.25)r[k+1]-0.75(0.25)r[k-1] + 0.25^2 r[k]
\end{split}
\end{equation}
which suggests that 
\begin{equation}
 0.375r[k] = \delta(k) + 1.3125\delta(k-1) - 0.25\delta(k-2) +0.1875(r[k+1] - r[k-1])
\end{equation}
For $k = 0$, we get
\begin{equation}
 0.375r[0] = \delta(0) + 1.3125\delta(-1) - 0.25\delta(-2) +0.1875(r[+1] - r[-1])
\end{equation}
we know that $\delta(0) = 1$ and $\delta(-1) = \delta(-2) = 0$. Also $r[1] = r[-1]$ due to the fact that $r[k]$ is even
\begin{equation}
 r[0] =\frac{1}{0.375} \delta(0) = \frac{8}{3}
\end{equation}
For $k = 1$, we get 
\begin{equation}
 0.375r[1] = 1.3125 + 0.1875(r[2] - r[0])
\end{equation}
which is (again because $r[k]$ is an even function)
\begin{equation}
 0.375r[1] = 1.3125 + 0.1875(r[2] - \frac{8}{3}) \tag{*}
\end{equation}
and at $k = -1$
\begin{equation}
 0.375r[-1] = 0.1875(r[0] - r[-2])
\end{equation}
which is (again because $r[k]$ is an even function)
\begin{equation}
 0.375r[1] = 0.1875(\frac{8}{3} - r[2])\tag{**}
\end{equation}
Now solve the system of two equations in two unknowns $(*),(**)$ to get $r[1],r[2]$, i.e.
\begin{equation}
 \begin{bmatrix}
    0.3750 &amp;  -0.1875\\
    0.3750  &amp;  0.1875  
 \end{bmatrix}
 \begin{bmatrix}
  r[1]\\
  r[2]
 \end{bmatrix}
 =
 \begin{bmatrix}
    0.8125\\
    0.5000
 \end{bmatrix}
\end{equation}