Need Help With These 2 Problems find k*x)=f(og) Need someone to check If I'am write I think I have the write solution. 
First question:

Given: $g(x)=2x^2+5$ and $f(x) = x^2+4x$, find $k(x) = (f\circ g)(\sqrt3)$.

My answer:

$k(x)=119+48\sqrt3$

Second question:

domain for $k(x)$ if: $d(x)=\sqrt{x-2}$ and $g(x) = 3\sqrt{4-x}$ and $k(x) = (d\circ g)(x)$ 

My answer:

domain = $\{x\mid 2\le x\le 4, x\in\Bbb R\}$

[ OP's original images, which I have translated into MathJax:
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 A: We have 
$$k(t)=(f\circ g)(t)=f(g(t)).$$
More informally, first do (apply)  $g$ to $t$, then do $f$ to the result.
Since $g(x)=2x^2+5$, we have $g(\sqrt{3})=11$. And now $f(11)=121+44$. So $k(\sqrt{3})=165$. 
Added: Now that the second question has been LaTeX'ed (thank you, MJD), it can be attacked.
We want the domain of $d\circ g$. To evaluate $(d\circ g)(x)$, first we find $g(x)$, and then apply $d$ to the result. The function $g$ is only defined if $x\le 4$. And the function $d$ is only defined for inputs that are $\ge 2$.
Recall that what gets fed into the function $d$ is $g(x)$, that is, $3\sqrt{4-x}$.  So we need $x\le 4$ and $3\sqrt{4-x}\ge 2$. 
For $x\le 4$, the condition $3\sqrt{4-x}\ge 2$ is equivalent to $9(4-x)\ge 4$. (We squared both sides.) This can be rearranged as $9x\le 32$, and then as $x\le \frac{32}{9}$. 
So for the function $k(x)$ to be defined, we want $x\le 4$ and $x\le \frac{32}{9}$. The condition $x\le 4$ has become superfluous, since it is included in the condition $x\le \frac{32}{9}$. So $k(x)$ is defined for all real numbers $x$ such that $x\le \frac{32}{9}$. 
Remark: Another version of the second question has $(dgg)(x)$, where presumably ordinary product is meant, since there is no little circle $\circ$.  If that is the intended function, then indeed the domain is the set of all $x$ such that $2\le x\le 4$.
