Answer
$ \displaystyle \frac{(x-3)^{2}}{2}+(y-4)^{2}=1$
New foci:$\quad (2, 4)$ and $(4, 4)$
New vertices:$\quad (3\pm\sqrt{2},4)$
New center:$\quad (3,4)$
Work Step by Step
Shift right by $3 \Rightarrow$ in the equation, replace $x$ with $x-3$
Shift up by $4 \Rightarrow$ in the equation, replace $y$ with $y-4$
New equation:$\displaystyle \quad \frac{(x-3)^{2}}{2}+(y-4)^{2}=1$
The ellipse $\displaystyle \frac{x^{2}}{2}+\frac{y^{2}}{1}=1$ is of the form:
Foci on the x-axis: $\quad \displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \quad (a\gt b)$
$a=\sqrt{2},\quad b=$1
Center-to-focus distance: $\quad c=\sqrt{a^{2}-b^{2}}=\sqrt{2-1}=1$
Foci: $\quad(\pm 1, 0)$
Vertices: $\quad (\pm\sqrt{2}, 0)$
The translations are such that $(x,y)\rightarrow(x',y')$, where
$\left\{\begin{array}{ll}
x'=x+3 & \text{... shift right}\\
y'=y+4 & \text{... shift up}
\end{array}\right.$
New foci:
$\quad (\pm 1+3, 0+4)\Rightarrow\quad (2, 4)$ and $(4, 4)$
New vertices:
$\quad (\pm\sqrt{2}+3, 0+4)\Rightarrow\quad (3\pm\sqrt{2},4)$
New center:
$\quad (0+3,0+4)\quad =(3,4)$
