I wasn't able to figure out what you did, so here is my analysis, without the resistance. Let:
Q = Total volume flow rate
$Q_a$ = Volume flow rate into converging pipe
$Q_b$ = Volume flow rate into diverging pipe
$p_1a$ = static pressure just after entrance to a
$p_2a$ = static pressure just before exit from a
$p_1b$ = static pressure just after entrance to a
$p_2b$ = static pressure just before exit from a
$T_1$ = "total pressure" in channel leading up to diffluence
$T_2$ = "total pressure" in channel after diffluence
$A_{a1}$ = cross sectional area of converging pipe at inlet
$A_{a2}$ = cross sectional area of converging pipe at outlet
$A_{b1}$ = cross sectional area of diverging pipe at inlet
$A_{b2}$ = cross sectional area of diverging pipe at outlet
CASE OF NO FRICTIONAL LOSS
Bernoulli equations relevant to pipe a:
$$T_1=p_1+\rho \frac{(Q_a/A_{a1})^2}{2}$$
$$p_1+\rho \frac{(Q_a/A_{a1})^2}{2}=p_2+\rho \frac{(Q_a/A_{a2})^2}{2}$$
$$p_2+\rho \frac{(Q_a/A_{a2})^2}{2}=T_2$$
Adding these three equations together gives $$T_1=T_2$$
Thus, for the case without friction, energy is conserved and the "total pressure" after the split section is equal to the "total pressure" before the split section. This is irrespective of how the flow splits between the two sections. The Bernoulli equations for pipe b will give the same result. Also, the convergence and divergence in the channels doesn't matter, as long as the final outlet pipe has the same cross sectional area as the initial inlet pipe.
CASE WITH FRICTIONAL EFFECTS INCLUDED
Bernoulli equations relevant to pipe a:
$$T_1=p_1+\rho \frac{(Q_a/A_{a1})^2}{2}$$
$$p_1+\rho \frac{(Q_a/A_{a1})^2}{2}=p_2+\rho \frac{(Q_a/A_{a2})^2}{2}+k_a\rho \frac{(Q_a/A_{a1})^2}{2}$$
$$p_2+\rho \frac{(Q_a/A_{a2})^2}{2}=T_2$$
Adding these three equations together gives $$T_1=T_2+k_a\rho \frac{(Q_a/A_{a1})^2}{2}\tag{1}$$
Similarly, for channel b:$$T_1=T_2+k_b\rho \frac{(Q_b/A_{b1})^2}{2}\tag{2}$$
Thus, for the case with friction, mechanical energy is not conserved and the "total pressure" after the split section is not equal to the "total pressure" before the split section. Moreover, the split between the two channels is relevant.
Mass balance equation:
$$Q_a+Q_b=Q\tag{3}$$
Eqns. 1-3 provide three algebraic equations in the three unknowns $(T_1-T_2)$, $Q_a$, and $Q_b$.
This is a correct way to solve exercises involving viscosity (within certain constraints, e.g., constant viscosity). Eqn. 1 is a version of the Bernoulli equation, modified to include a frictional head loss, and is definitely valid, provided the velocities used are the average velocities. Eqn. 1 without the $h_L$ is valid along a streamline, even for a viscous flow. If Eqn. 1 is being used for a laminar viscous flow (say in a tube of slowly varying cross section), the kinetic energy terms should not have a 2 in the denominator. See Bird, Stewart, and Lightfoot, Transport Phenomena for details. Here they show that, if the 2 is to be included, then instead of using the average value of v squared, one should use the average value of $v^3$ divided by the average value of v. For laminar flow in a tube, this reduces to twice the square of the average velocity, so, if the average velocity squared is being used in the kinetic energy term, the 2 should not be included in the denominator if the flow is laminar.
Best Answer
The head loss is the requiring pressure to create a given flow. The head loss will be the same for the tree pipes (if we neglect potential difference due to gravity and pipe height) since it is set by pressure difference between tank A and B. But flows through the tree pipes will be different. For a given head loss, at constant friction factor, flow will be greater for larger pipe diameter.
Edit:
The equation you cited is called the Darcy–Weisbach equation. It is a phenomenological law (such as Ohm Law) where the friction factor (Darcy factor is dependent of the flow). It can be assessed using Moody diagram (friction factor versus Reynolds number) or dimensionless formula.
To overcome your difficulty you must realize that: