Sin2 42°+sin248°+sec2 30°-cot2 45°

We know that $\sin(90^\circ - \theta) = \cos \theta$.
So, $\sin 48^\circ = \sin(90^\circ - 42^\circ) = \cos 42^\circ$.
Therefore, the first two terms become:

$$\sin^2 42^\circ + \sin^2 48^\circ = \sin^2 42^\circ + \cos^2 42^\circ$$

Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$:

$$\sin^2 42^\circ + \cos^2 42^\circ = 1$$

We know that $\cos 30^\circ = \frac{\sqrt{3}}{2}$.
Since $\sec \theta = \frac{1}{\cos \theta}$, we have $\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$.
Therefore:

$$\sec^2 30^\circ = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}$$

We know that $\tan 45^\circ = 1$.
Since $\cot \theta = \frac{1}{\tan \theta}$, we have $\cot 45^\circ = \frac{1}{\tan 45^\circ} = \frac{1}{1} = 1$.
Therefore:

$$\cot^2 45^\circ = 1^2 = 1$$

The original expression is $\sin^2 42^\circ + \sin^2 48^\circ + \sec^2 30^\circ - \cot^2 45^\circ$.
Substituting the values from Step 1, Step 2, and Step 3:

$$1 + \frac{4}{3} - 1$$

$$1 + \frac{4}{3} - 1 = \frac{4}{3}$$

The final answer is $\frac{4}{3}$.