The ellipse is
$$x^2+4y^2=4\iff \frac{x^2}{2^2}+y^2=1$$
and the line is $\,y=-x+4\,$ , which is then "above" the ellipse all the time and, in particular, in the first quadrant (draw an approximate sketck of the functions to see why is this relevant).
Thus, we can write down a point on the ellipse in the first quadrant as
$$\left(x\,,\,\frac{1}{2}\sqrt{4-x^2}\right)\;,\;\;x\ge 0$$
so using the formula for the distance between a point as above and the line $\,x+y-4=0\,$ we get
$$\frac{\left|\;x+\frac{1}{2}\sqrt{4-x^2}-4\;\right|}{\sqrt2}=\frac{\left|\;2x+\sqrt{4-x^2}-8\;\right|}{2\sqrt2}$$
But since the expression with the absolute value is always negative (in $\,0\le x\le 2\,$ , that is)(why?) , we must get the minimum of:
$$f(x)=\frac{-2x-\sqrt{4-x^2}+8}{2\sqrt 2}\implies f'(x)=\frac{-2+\frac{x}{\sqrt{4-x^2}}}{2\sqrt 2}=0\iff$$
$$-2\sqrt{4-x^2}+x=0\iff x^2=16-4x^2\iff x=\frac{4}{\sqrt 5}$$
Check the above indeed is a minimum point and input in the resp. equation...
Of course, the maximal distance is...
$\newcommand{\minimize}{\operatorname*{Minimize}}$You can use the penalty method or the augmented Lagrangian method.
Suppose you have a problem of the form
\begin{aligned}
\mathbb{P}:\minimize_{g(x) \leq 0} f(x).
\end{aligned}
You can solve the sequence of
\begin{aligned}
\mathbb{P}_\lambda:\minimize_{x\in\mathbb{R}^n} f(x) + c \left[g(x)\right]_+^2,
\end{aligned}
where $\left[z\right]_+$ is defined as
\begin{aligned}
\left[z\right]_+ = \begin{cases}
0, &\text{if } z < 0
\\
z, &\text{if } z \geq 0
\end{cases}
\end{aligned}
In doing so, you will solve $\mathbb{P}_{\lambda_k}$, increase $\lambda_k$ (e.g., $\lambda_{k+1} = 10 \lambda_k$) and solve the next problem, $\mathbb{P}_{\lambda_k}$, using the previous solution as an initial guess.
Let us denote the set of minimizers of $\mathbb{P}_{\lambda_k}$ by $X_k^\star$ and let $X^\star$ be the set of minimizers of $\mathbb{P}$. Then, it can be shown that
$$
\limsup_k X_k^\star \subseteq X^{\star},
$$
in other words, every cluster point, $x^\star$, of a sequence $(x_k^\star)_k$ with $x_k^\star \in X_k^\star$, is a minimizer of $\mathbb{P}$.
Although in theory you need to take $\lambda$ to infinity in order to recover a solution of the original problem, in practice you can stop the procedure once $[g(x_k^\star)]_+$ drops below a desired tolerance.
Note that there exist penalty functions other than $c \left[g(x)\right]_+^2$ (for example, $c \left[g(x)\right]_+$). You can read more about the penalty method in the book of Nocedal and Wright.
The augmented Lagrangian method is similar, but it uses a different unconstrained formulation.
Best Answer
Because $a$ is a positive constant we can ignore it and minimise $$ F(x)=3 x_1+5 x_2 $$
Rewrite the inequalities with slack variables:
$$ \frac{2.16}{x_1}+\frac{10}{x_2}\leq 1 \implies \frac{2.16}{x_1}+\frac{10}{x_2}+s_1^2= 1\\ g_1=\frac{2.16}{x_1}+\frac{10}{x_2}+s_1^2-1 $$ Now the second one: $$ x_1 \geq 11.25\implies x_1 = 11.25+s_2^2\\ g_2= x_1-11.25-s_2^2 $$ Finally, the third one: $$ x_2 \geq 18.75 \implies x_2 = 18.75+s_3^2\\ g_3 = x_2 - 18.75-s_3^2 $$ Form the Lagrangian:
$$ \begin{split} L(x_1,x_2,\lambda_1,\lambda_2,\lambda_3,s_1,s_2,s_3) &= 3 x_1 + 5 x_2 \\ &- \lambda_1 \left(\frac{2.16}{x_1}+\frac{10}{x_2}+s_1^2-1\right) \\ &- \lambda_2 \left(x_1-11.25-s_2^2\right) \\ &- \lambda_3 \left(x_2 - 18.75-s_3^2\right) \end{split} $$ This implies the following partial derivatives: $$ \begin{split} \frac{\partial L}{\partial x_1} &= 3+\frac{2.16\lambda_1}{x_1^2}-\lambda_2\\ \frac{\partial L}{\partial x_2} &= 5+{{10\lambda_1}\over{x_2^2}}-\lambda_3\\ \frac{\partial L}{\partial s_1} &= -2s_1\lambda_1\\ \frac{\partial L}{\partial s_2} &= 2s_2\lambda_2\\ \frac{\partial L}{\partial s_2} &= 2s_3\lambda_3 \end{split} $$