1. Check out the wiki at http://hades.mech.northwestern.edu/wiki/index.php/Operational_Amplifiers_(Op-Amps).
a. You could use one op-amp as an inverting amplifier with a gain of -1 for Vref, then a second op-amp as an inverting summer.
b. Use an inverting amplifier with a gain of -10.
c. Use an op-amp as an integrator.
d. Use an op-amp as a differentiator.
e. Use an op-amp as a summer to sum (and take the negative of) the three signals.

It is a good idea to use resistors in the 1k - 100k range or so, so the resistances are comfortably less than the input impedance of the op-amps but not so low as to be wasting power in your circuit or taxing the current supply capability of the sources of Vsens and Vref.

2. Current I1 flows from V1 through R1 and R to Vout. Current I2 flows from V2 to ground through R2 and R. No current enters the op-amp. Because there is negative feedback, V+ = V- (the voltages at the two op-amp inputs). We have
V+ = V2 (R/(R+R2))
so
I1 = (V1 - (V2 (R/(R+R2))))/R1
and
Vout = V2 (R/(R+R2)) - I1*R
which you can then simplify.

3. This is a rectifier. When Vin drops to -0.7 V or lower, the diode is forward biased and current flows. Vout can never go less than -0.7 V, because a diode never allows more than 0.7 V across it in the forward direction. When Vin is greater than -0.7 V, the diode is not forward biased, and no current flows, so Vout = Vin. So the output voltage equals the input voltage, except when the input voltage is less than -0.7 V; then, the output voltage is -0.7 V.

4. The transistor comes on when the base voltage Vb > 0.7 V, so the transistor is off for Vin < 1.4 V (due to the voltage divider made by the two 1k resistors). The transistor is saturated when the collector voltage Vc is 9.3 V due to the 0.7 V diode drop, and Vce = 0.2 V, so Ve = 9.1 V. That means there must be an emitter current of (9.1 V)/(100 ohms) = 0.091 A = 91 mA flowing through the 100 ohm resistor. Ie = Ib + Ic = Ib + 100*Ib = 101 Ib, so Ib = 0.9 mA. Vb = 9.1 V + 0.7 V (diode drop) = 9.8 V. The current Iin coming from Vin must be Ib + (9.8 V)/(1000 ohm), since some of the current goes into the base and some flows through the 1k resistor to ground. So we get
Iin = 0.0009 A + 0.0098 A = 0.0107 A
and
Vin = 9.8 V + Iin*(1000 ohms) = 20.5 V.
So the LED is off for Vin < 1.4 V and at its maximum brightness for Vin > 20.5 V.