Expand the square of the sum:

\((\sin\alpha+\cos\alpha)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha\).

Since \(\sin^2\alpha+\cos^2\alpha=1\) and \(2\sin\alpha\cos\alpha=\sin2\alpha\), we get

\((\sin\alpha+\cos\alpha)^2=1+\sin2\alpha\).

For \(\alpha=-15^\circ\):

\(1+\sin(-30^\circ)=1-\dfrac12=\dfrac12\).

Use the definition of tangent:

\(\tan^2\alpha=\dfrac{\sin^2\alpha}{\cos^2\alpha}\).

Then

\(\tan^2\alpha\cdot2\cos^2\alpha=2\sin^2\alpha\).

Therefore, the whole expression equals

\(2-2\sin^2\alpha=2(1-\sin^2\alpha)=2\cos^2\alpha\).

For \(\alpha=-\dfrac{\pi}{4}\):

\(2\cos^2\left(-\dfrac{\pi}{4}\right)=2\cdot\dfrac12=1\).

Use the double-angle formula:

\(\sin4\alpha=2\sin2\alpha\cos2\alpha\).

Then

\(\dfrac{\sin4\alpha}{\cos2\alpha}=2\sin2\alpha\).

For \(\alpha=15^\circ\):

\(2\sin30^\circ=2\cdot\dfrac12=1\).

Transform the expression in parentheses:

\(\cos2\alpha+\sin^2\alpha=(\cos^2\alpha-\sin^2\alpha)+\sin^2\alpha=\cos^2\alpha\).

Then the whole expression becomes

\(6\cos^2\alpha\tan^2\alpha=6\cos^2\alpha\cdot\dfrac{\sin^2\alpha}{\cos^2\alpha}=6\sin^2\alpha\).

For \(\alpha=\dfrac{3\pi}{4}\):

\(6\sin^2\dfrac{3\pi}{4}=6\cdot\dfrac12=3\).

Use the reduction formulas:

\(\tan(\pi-\alpha)=-\tan\alpha\), \(\sin\left(\dfrac{\pi}{2}-\alpha\right)=\cos\alpha\).

Then the expression becomes

\(4(-\tan\alpha)\cos\alpha=-4\dfrac{\sin\alpha}{\cos\alpha}\cos\alpha=-4\sin\alpha\).

For \(\alpha=\dfrac{5\pi}{6}\):

\(-4\sin\dfrac{5\pi}{6}=-4\cdot\dfrac12=-2\).

Use the reduction formulas:

\(\sin(\pi+\alpha)=-\sin\alpha\), \(\sin\left(\dfrac{\pi}{2}+\alpha\right)=\cos\alpha\).

Then

\(4\sqrt2\sin(\pi+\alpha)\sin\left(\dfrac{\pi}{2}+\alpha\right)=4\sqrt2(-\sin\alpha)\cos\alpha=-2\sqrt2\sin2\alpha\).

For \(\alpha=\dfrac{\pi}{8}\):

\(-2\sqrt2\sin\dfrac{\pi}{4}=-2\sqrt2\cdot\dfrac{\sqrt2}{2}=-2\).

Use the formulas:

\(1-\cos2\alpha=2\sin^2\alpha\), \(\cos\left(\dfrac{\pi}{2}-\alpha\right)=\sin\alpha\).

Therefore,

\(\dfrac{1-\cos2\alpha}{\cos\left(\dfrac{\pi}{2}-\alpha\right)}=\dfrac{2\sin^2\alpha}{\sin\alpha}=2\sin\alpha\).

For \(\alpha=150^\circ\):

\(2\sin150^\circ=2\cdot\dfrac12=1\).

Using the reduction formula, \(\cos(90^\circ+\alpha)=-\sin\alpha\), hence

\(\cos^2(90^\circ+\alpha)=\sin^2\alpha\).

Then

\(10\cos^2\alpha-10\sin^2\alpha=10(\cos^2\alpha-\sin^2\alpha)=10\cos2\alpha\).

For \(\alpha=\dfrac{\pi}{6}\):

\(10\cos\dfrac{\pi}{3}=10\cdot\dfrac12=5\).

Expand the square of the difference:

\((\sin\alpha-\cos\alpha)^2=\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha\).

Since \(\sin^2\alpha+\cos^2\alpha=1\) and \(2\sin\alpha\cos\alpha=\sin2\alpha\), we obtain

\((\sin\alpha-\cos\alpha)^2=1-\sin2\alpha\).

For \(\alpha=75^\circ\):

\(1-\sin150^\circ=1-\dfrac12=\dfrac12\).

Use the definition of cotangent:

\(\cot^2\alpha=\dfrac{\cos^2\alpha}{\sin^2\alpha}\).

Then

\(\cot^2\alpha\cdot\sin^2\alpha=\cos^2\alpha\).

Therefore, the expression equals

\(3-3\cos^2\alpha=3(1-\cos^2\alpha)=3\sin^2\alpha\).

For \(\alpha=\dfrac{3\pi}{4}\):

\(3\sin^2\dfrac{3\pi}{4}=3\cdot\dfrac12=\dfrac32\).

Use the double-angle formula:

\(\sin6\alpha=2\sin3\alpha\cos3\alpha\).

Then

\(\dfrac{\sin6\alpha}{\sin3\alpha}=2\cos3\alpha\).

For \(\alpha=20^\circ\):

\(2\cos60^\circ=2\cdot\dfrac12=1\).

Transform the expression in parentheses:

\(\cos2\alpha-\cos^2\alpha=(\cos^2\alpha-\sin^2\alpha)-\cos^2\alpha=-\sin^2\alpha\).

Then the whole expression equals

\(-12\sin^2\alpha\cot^2\alpha=-12\sin^2\alpha\cdot\dfrac{\cos^2\alpha}{\sin^2\alpha}=-12\cos^2\alpha\).

For \(\alpha=\dfrac{5\pi}{4}\):

\(-12\cos^2\dfrac{5\pi}{4}=-12\cdot\dfrac12=-6\).

Using the reduction formulas:

\(\cot(90^\circ-\alpha)=\tan\alpha\), \(\cos(180^\circ-\alpha)=-\cos\alpha\).

Then

\(4\cot(90^\circ-\alpha)\cos(180^\circ-\alpha)=4\tan\alpha\cdot(-\cos\alpha)=-4\sin\alpha\).

For \(\alpha=\dfrac{\pi}{6}\):

\(-4\sin\dfrac{\pi}{6}=-4\cdot\dfrac12=-2\).

Using the reduction formulas:

\(\cos(\pi-\alpha)=-\cos\alpha\), \(\cos\left(\dfrac{\pi}{2}+\alpha\right)=-\sin\alpha\).

Then

\(6\sqrt2\cos(\pi-\alpha)\cos\left(\dfrac{\pi}{2}+\alpha\right)=6\sqrt2\sin\alpha\cos\alpha=3\sqrt2\sin2\alpha\).

For \(\alpha=-\dfrac{\pi}{8}\):

\(3\sqrt2\sin\left(-\dfrac{\pi}{4}\right)=3\sqrt2\cdot\left(-\dfrac{\sqrt2}{2}\right)=-3\).

Use the formulas:

\(1+\cos2\alpha=2\cos^2\alpha\), \(\sin\left(\dfrac{\pi}{2}-\alpha\right)=\cos\alpha\).

Therefore,

\(\dfrac{1+\cos2\alpha}{\sin\left(\dfrac{\pi}{2}-\alpha\right)}=\dfrac{2\cos^2\alpha}{\cos\alpha}=2\cos\alpha\).

For \(\alpha=120^\circ\):

\(2\cos120^\circ=2\cdot\left(-\dfrac12\right)=-1\).

Using the reduction formula:

\(\cos(270^\circ-\alpha)=-\sin\alpha\).

Hence \(\cos^2(270^\circ-\alpha)=\sin^2\alpha\).

Then

\(8\cos^2\alpha-8\sin^2\alpha=8(\cos^2\alpha-\sin^2\alpha)=8\cos2\alpha\).

For \(\alpha=\dfrac{\pi}{4}\):

\(8\cos\dfrac{\pi}{2}=0\).