Question

Three conducting plates, each of area A, are connected as shown.

Answer 1

You have effectively got two capacitors in parallel. The effective capacitance is just the sum of the two. 
Cequiv = ε₀A/d₁ + ε₀A/d₂ Take these over a common denominator (d₁d₂) 
Cequiv = ε₀d₂A + ε₀d₁A / (d₁d₂) Cequiv = ε₀A( (d₁ + d₂) / (d₁d₂) ) 
B) It's tempting to just wave your arms and say that when d₁ or d₂ tends to zero C -> ∞, so the minimum will occur in the middle, where d₁ = d₂ 
But I suppose we ought to kick that idea around a bit. 
(d₁ + d₂) is effectively a constant. It's the distance between the two outer plates. Call it D. 
C = ε₀AD / d₁d₂ We can also say: d₂ = D - d₁ C = ε₀AD / d₁(D - d₁) C = ε₀AD / d₁D - d₁² 
Differentiate with respect to d₁ 
dC/dd₁ = -ε₀AD(D - 2d₁) / (d₁D - d₁²)² {d2C/dd₁² is positive so it will give us a minimum} For max or min equate to zero. 
-ε₀AD(D - 2d₁) / (d₁D - d₁²)² = 0 -ε₀AD(D - 2d₁) = 0 ε₀, A, and D are all non-zero, so (D - 2d₁) = 0 d₁ = ½D 
In other words when the middle plate is halfway between the two outer plates, (quelle surprise) so that 
d₁ = d₂ = ½D so 
Cmin = ε₀AD / (½D)² Cmin = 4ε₀A / D Cmin = 4ε₀A / (d₁ + d₂)