$Xt = 5+0.8(X_{t−1} −5)+0.8(X_{t−2} −5)−0.6(X_{t−3} −5)−0.5e_{t−1} + 0.25e_{t−2} +e_t$
Write the process using characteristic equations then determine ARIMA order and E[Xt]. Finally, find the values for E [▽Xt] and determine if ▽Xt is stationary and invertible. Actually, I am confused how to find E[Xt].
Observe that you may expand the equation as follows \begin{aligned} X_t = 5 + 0.8 X_{t-1}- 4 + 0.8 X_{t-2} - 4 - 0.6 X_{t-3} + 3 + e_t - 0.5 e_{t-1} + 0.25 e_{t-2} \end{aligned} which is the same as \begin{aligned} X_t - 0.8 X_{t-1} - 0.8 X_{t-2} + 0.6 X_{t-3} = e_t - 0.5 e_{t-1} + 0.25 e_{t-2} \end{aligned} Using the backward shift operator $q^{-1}$ (defined by $q^{-k} X_t = X_{t-k}$ for every $t$ and positive $k$), it follows that \begin{aligned} (1 - 0.8 q^{-1}- 0.8 q^{-2} + 0.6 q^{-3}) X_t = (1 - 0.5 q^{-1} + 0.25 q^{-2})e_t \end{aligned}
This gives that \begin{aligned} X_t = \frac{(1 - 0.5 q^{-1} + 0.25 q^{-2})}{(1 - 0.8 q^{-1}- 0.8 q^{-2} + 0.6 q^{-3}) }e_t \end{aligned} assuming that the division by $(1 - 0.8 q^{-1}- 0.8 q^{-1} + 0.6 q^{-1}) $ is valid.
Now you can compute the expectation. For example, $E[X_t] = 0$ for all $t$ if the errors $e_t$ have zero mean for all $t$.
The stationary of $X_t$ depends on the stationary of $e_t$ and the stability of the model. You need to check the roots of the characteristic polynomial (the denominator) and check the location of the roots with respect to the unit circle. It is easy to see that $X_t$ has a unit root and therefore it is not stationary.
From the second equation above one can see that (by multiplying both sides by the backward shift operator) \begin{aligned} X_{t-1} - 0.8 X_{t-2} - 0.8 X_{t-3} + 0.6 X_{t-4} = e_{t-1} - 0.5 e_{t-2} + 0.25 e_{t-3} \end{aligned}
This means that \begin{aligned} (X_t - X_{t-1}) &- 0.8 (X_{t-1} - X_{t-2}) - 0.8 (X_{t-2}-X_{t-3}) + 0.6 (X_{t-3} - X_{t-4}) \\ &= e_t - 1.5 e_{t-1} + 0.75 e_{t-2} -0.25 e_{t-3} \end{aligned} which is the same as \begin{aligned} \nabla X_t &- 0.8 \nabla X_{t-1} - 0.8 \nabla X_{t-2} + 0.6 \nabla X_{t-3} = e_t - 1.5 e_{t-1} + 0.75 e_{t-2} -0.25 e_{t-3} \end{aligned} and \begin{aligned} \nabla X_t =\frac{(1 - 1.5 q^{-1} + 0.75 q^{-2} - 0.25q^{-3})}{(1 - 0.8 q^{-1}- 0.8 q^{-2} + 0.6 q^{-3}) }e_t \end{aligned}
Therefore, $E[\nabla X_t] = 0$ for all $t$ if $E[e_t] = 0$ for all $t$, and stationary is checked again by checking the roots of the polynomial $$ 1 - 0.8 q^{-1}- 0.8 q^{-2} + 0.6 q^{-3} $$ in a similar manner as for $X_t$ above. It is easy to see that $\nabla X_t$ has a unit root and therefore it is not stationary. But because the roots of the numerator are strictly inside the unit circle, the process is invertible.