Stepping down voltage
I need a current of 3.5A to drive a LOAD(a Nichrome wire of length 5-6cm) and 7.4V. How do i proceed with this ?


Let's get started with a couple of equations that we're going to need.

Sloppy Ohm's law: \$V=A×Ω\$

Sloppy power equation: \$W = V×I = \frac{V^2}{R}\$

By saying sloppy I mean that I'm using the units rather than their common variables that holds their values (U for voltage, I for ampere etc..). 



The power dissipated in the nichrome wire is \$7.4V×3.5A=25.9W\$

So let's see how much current we would only require if we want \$25.9W\$ and we're using \$11.1V\$ instead.

\$W=V×A &lt;=&gt; A = \frac{W}{V} = \frac{25.9W}{11.1V}\approx 2.3A\$

Now how are you going to get 2.3A instead of 3.5A? Either use a buck converter that is 100% efficient, or do something else that's much easier and 100% efficient (no step down loss). 

Just make the nichrome wire longer, that way you will linearly reduce the power dissipated into the nichrome wire. It is only linearly reduced because the voltage is fixed and I'm treating your battery as an ideal voltage source. 

So we know that \$W=\frac{V^2}{R}\$, we know that you want \$25.9W\$, let's calculate how much you're getting with the "5-6" cm you got now. 

The resistance of the nichrome wire is \$\frac{7.4V}{3.5A}\approx 2.1Ω\$

\$W = \frac{11.1V^2}{2.1Ω} \approx 58W\$, this means that you need a wire that is \$\frac{58W}{25.9W}=2.25\$ times longer. So if the wire is 5 cm, then you can replace it with \$11.25\$ cm wire and get \$25.9W\$ dissipated in your new longer nichrome wire. 

And there's no transistor involved anywhere, no step down losses, the only real losses is that first nichrome wire you got that you might have to throw away.