MOS differential amplifier question please help
Here are some hints.

I trust that M1 and M2 are actually both NMOS and that there's a mistake in the schematic drawing. 

You should know (from lectures or nodal analysis) that: \$I_{ds1} + I_{ds2} = I_T\$ 

(a) 

The differential output voltage equal to zero implies V0+ = V0-

Thus this must be true: \$I_{RD1} = I_{RD2}\$

And therefore also: \$I_{ds1} = I_{ds2} = 0.5 I_T = 0.2\$mA

So for M1 we have: \$24\frac{µ_nC_{ox}}{2}(V_{gs1}-0.7)^2 = 0.2\$mA

The equation for M2 is similar.

Solve for the two \$V_{gs}\$ and find their difference.

(b) 

Remember: \$I_{ds1} + I_{ds2} = I_T\$

So if \$I_{ds1} = 0.7 I_T\$ then you must have \$I_{ds2} = 0.3 I_T\$.

The rest of the solution can be done as in (a), just change the currents.

(c)

Since the load is resistive the output common-mode will change linearly with \$I_T\$

Which node changes with Vcm_in and how does that affect Vcm_out ?

I think you can assume, for simplicity, that Ids1=Ids2 and both Vgs1 and Vgs2 are constant. 

(d)

Here you should use small-signal analysis. First, you need to find estimates of the differential gain and then the common-mode gain.

It also seems reasonable to assume \$r_{ds1} \gg R_{D1}\$ and \$r_{ds2} \gg R_{D2}\$ then you can make simplifications like \$g_{m1}(r_{ds1} \parallel R_{D1}) \approx g_{m1}R_{D1}\$