To solve the problem, you need to know the following information:

$\displaystyle sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

$\displaystyle tan(45^{\circ})=1$

$\displaystyle sin(45^{\circ})=\frac{\sqrt{2}}{2}$

Replace the trigonometric functions with these values:

$\displaystyle \frac{4sin^2(60^{\circ})-tan(45^{\circ})}{3sin(45^{\circ})}$

$\displaystyle \frac{4(\frac{\sqrt{3}}{2})^2-(1)}{3(\frac{\sqrt{2}}{2})}$

$\displaystyle \frac{4(\frac{3}{4})-(1)}{(\frac{3\sqrt{2}}{2})}$

$\displaystyle \frac{2}{(\frac{3\sqrt{2}}{2})}$

$\displaystyle \frac{4}{3\sqrt{2}} = \frac{4\cdot \sqrt{2}}{3\sqrt{2}\cdot \sqrt{2}} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}$

To determine the value of the expression, you must know the following trigonometric values:

$\displaystyle cos(45^{\circ}) = \frac{\sqrt{2}}{2}$

$\displaystyle sin (30^{\circ})=\frac{1}{2}$

Replacing these values, we get:

$\displaystyle cos^{2}(45^{\circ})-sin(30^{\circ})$

$\displaystyle (\frac{\sqrt{2}}{2})^{2}-\frac{1}{2}$

$\displaystyle \frac{2}{4}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0$

$\displaystyle tan (x)*sin(x) + sec(x)*(cos x)^{2} =$

$\displaystyle \frac{sin (x)}{cos (x)}* sin(x) + \frac{1}{cos (x)}*(cos x)^{2} =$

$\displaystyle \frac{(sin x)^{2} + (cos x)^{2}}{cos (x)} = \frac{1}{cos (x)} = sec (x)$

We need to use the following identities:

$\displaystyle cos(\frac{3\pi}{2}+x)=sin\ x$

$\displaystyle sin (\pi+x)=-sin\ x$

$\displaystyle cos(\frac{\pi}{2}+x)=-sin\ x$

$\displaystyle sin(\pi-x)=sin\ x$

Use these to simplify the expression as follows:

$\displaystyle A=cos(\frac{3\pi}{2}+x)+sin (\pi+x)+cos(\frac{\pi}{2}+x)+sin(\pi-x)$

$\displaystyle \Rightarrow A=sin\ x-sin\ x-sin\ x+sin\ x = 0$

$\displaystyle sin(\frac{\pi}{2})=1$

$\displaystyle cos(\frac{\pi}{2})=0$

Plug these values in:

$\displaystyle A=\frac{2sin(\frac{\pi}{2})+4cos(\frac{\pi}{2})}{4sin(\frac{\pi}{2})-cos(\frac{\pi}{2})}=\frac{2\times 1+4\times 0}{4\times 1-0}=\frac{2}{4}=\frac{1}{2}=0.5$

$\displaystyle tan\ x=\frac{1}{4}\Rightarrow \frac{sin\ x}{cos\ x}=\frac{1}{4}\Rightarrow cos\ x=4sin\ x$

Substitute $\displaystyle cos\ x=4sin \ x$ into the expression:

$\displaystyle D=\frac{sin\ x}{sin\ x-cos\ x}+\frac{sin\ x+cos\ x}{cos\ x}=\frac{sin\ x}{sin\ x-4sin\ x}+\frac{sin\ x+4sin\ x}{4sin\ x}$

$\displaystyle \Rightarrow D=\frac{sin\ x}{-3sin\ x}+\frac{5sin\ x}{4sin\ x}$

$\displaystyle \Rightarrow D=-\frac{1}{3}+\frac{5}{4}=\frac{-4+15}{12}=\frac{11}{12}$

 $\displaystyle cot\ x=3\Rightarrow \frac{cos\ x}{sin\ x}=3\Rightarrow cos\ x= 3sin\ x$

Now substitute $\displaystyle cos\ x=3sin\ x$ into the expression:

$\displaystyle D=\frac{cos^2x}{(sin\ x)(cos\ x)}-\frac{sin\ x+cos\ x}{sin\ x}=\frac{(3sin\ x)^2}{(sin\ x)(3sin\ x)}-\frac{sin\ x+3sin\ x}{sin\ x}$

$\displaystyle \Rightarrow D=\frac{9sin^2x}{3sin^2x}-\frac{4sin\ x}{sin\ x}$

$\displaystyle \Rightarrow D=\frac{9}{3}-4=3-4=-1$

We need to use the following identitities:

$\displaystyle \frac{1}{cos ^2x}=1+tan^2x$

$\displaystyle \frac{1}{sin^2x}=1+cot^2x$

Now substitute them into the expression:

$\displaystyle A=\frac{sin^2x}{1+tan^2x}-\frac{cos^2x}{1+cot^2x}+1=\frac{sin^2x}{\frac{1}{cos^2x}}-\frac{cos^2x}{\frac{1}{sin^2x}}+1$

$\displaystyle \Rightarrow A=(sin^2x)(cos^2x)-(sin^2x)(cos^2x)+1=0+1=1$

Based on the double angle formula we have, $\displaystyle sin\ 2x=2(sin\ x)(cos\ x)$.

$\displaystyle \frac{sin\ x+sin\ 2x}{sin\ x-sin\ 2x}=\frac{sin\ x+2(sin\ x)(cos\ x)}{sin\ x-2(sin\ x)(cos\ x)}$

$\displaystyle =\frac{sinx(1+2cos\ x )}{sinx(1-2cos\ x )}=\frac{1+2cos\ x}{1-2cos\ x}$

$\displaystyle =\frac{1+2(\frac{1}{3})}{1-2(\frac{1}{3})}=\frac{\frac{3+2}{3}}{\frac{3-2}{3}}=5$

$\displaystyle x=\frac{\pi}{3}\Rightarrow sin\ x=sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$

Now we can write:

$\displaystyle x=\frac{\pi}{3}\Rightarrow 2x=2(\frac{\pi}{3})=\frac{2\pi}{3}\Rightarrow cos\ 2x=cos(\frac{2\pi}{3})$

$\displaystyle =cos(\pi-\frac{\pi}{3})$

$\displaystyle =-cos (\frac{\pi}{3})$

$\displaystyle =-\frac{1}{2}$

Now we can substitute the values:

$\displaystyle \frac{sin\ x}{sin\ x+cos\ 2x}=\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}-\frac{1}{2}}=\frac{\sqrt{3}}{\sqrt{3}-1}=\frac{\sqrt{3}}{\sqrt{3}-1}\times \frac{{\sqrt{3}+1}}{{\sqrt{3}+1}}=\frac{3+\sqrt{3}}{2}$